Find and measure angle ACB ​

It is not required to calculate the standard error of a sample mean or the standard error of an estimate in general.

False. The standard error is a measure of the variability of the sample mean and is calculated using the sample standard deviation and sample size, without necessarily requiring the calculation of the pooled variance.

The pooled variance, on the other hand, is a statistic used in hypothesis testing when comparing means from two independent samples, assuming that the two populations have equal variances. It is calculated by pooling the variances of the two samples, weighted by their degrees of freedom, and is used to calculate the standard error of the difference between the means.

While the pooled variance can be used to calculate the standard error of the difference between two means, it is not required to calculate the standard error of a sample mean or the standard error of an estimate in general.

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By squaring the deviations from the mean, we make them positive numbers, and the sum of the squared deviations will also be a positive number.

In mathematics and statistics, deviation refers to the difference between a value and a reference value or an expected value. More specifically, deviation is a measure of how far a set of numbers is spread out from their average value or the central point.

By squaring the deviations from the mean, we make them positive numbers, and the sum of the squared deviations will also be a positive number. This is because squaring any number makes it positive, regardless of whether the original number was positive or negative.

Additionally, squaring the deviations before summing them up allows us to give more weight to larger deviations from the mean. This is because squaring a larger deviation will result in a much larger value than squaring a smaller deviation, which helps to highlight the effect of outliers or extreme values in the data set.

The sum of the squared deviations is used in many statistical calculations, including calculating the variance and standard deviation of a data set, which are measures of the spread of the data.

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The mean slope of the function on the interval [-6, 10] is:

Mean slope is -62.

To find the mean slope of the function f(x) on the interval [-6, 10], we

need to calculate the total change in the function over that interval and

divide by the length of the interval:

Mean slope = (f(10) - f(-6)) / (10 - (-6))

We can simplify this by first finding the derivative of the function f(x):

[tex]f'(x) = 6x^2 - 18x - 108[/tex]

Then, we can use the mean value theorem to find the two values of c

where the instantaneous slope of the function equals the mean slope:

[tex]c1 = (-6 + \sqrt{(783)} ) / 3\\c2 = (-6 - \sqrt{ (783)} ) / 3[/tex]

Plugging these values into the derivative of f(x) gives the instantaneous

slope at each value of c:

f'(c1) = 165

f'(c2) = -33

Therefore, the mean slope of the function on the interval [-6, 10] is:

Mean slope = (f(10) - f(-6)) / (10 - (-6)) = (57 - 1015) / 16 = -62

And we can conclude that the mean slope of the function f(x) on the

interval [-6, 10] is equal to -62.

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The null hypothesis (H0) is that there is no difference in the average time taken to read a book before and after the program.

A middle school teacher conducts a study to evaluate the impact of a reading rewards program on the time taken by students to read a book. To analyze the data, the teacher will use a two-sample t-test. The population parameters being compared are the average time taken to read a book before and after the program.

The level of significance is a threshold value that determines the probability of making a Type I error (rejecting a true null hypothesis). It is usually denoted by  (alpha) and is typically set at 0.05, which means there is a 5% chance of making a type I error. However, the specific level of significance will depend on the teacher's chosen value.

In this case, the t-test should be a paired t-test because the same group of students is being compared before and after the program, making the two samples dependent on each other.

The null hypothesis (H0) is that there is no difference in the average time taken to read a book before and after the program. In other words, the average difference between the pre- and post-program reading times is equal to zero.

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Logarithmic differentiation to evaluate f′(x)f(x)=(3x)ln3xf′(x) will give f′(x) = 3ln3x + 3/x.

To use logarithmic differentiation to evaluate f′(x) for the given function f(x)=(3x)ln3x, we first take the natural logarithm of both sides:
ln(f(x)) = ln[(3x)ln3x]
Using the properties of logarithmic, we can simplify this expression:
ln(f(x)) = ln(3x) + ln(ln3x)
ln(f(x)) = ln(3) + ln(x) + ln(ln3) + ln(x)
Now we differentiate both sides of this equation with respect to x, using the chain rule on the right-hand side:
(1/f(x))f′(x) = 1/x + 1/x ln3 + 1/ln3 + 1/x
Simplifying this expression, we get:
f′(x) = f(x) [(1/x) + (1/x ln3) + (1/ln3) + (1/x)]
Substituting in the original expression for f(x), we get:
f′(x) = (3x)ln3x [(1/x) + (1/x ln3) + (1/ln3) + (1/x)]
Simplifying this expression, we get:
f′(x) = 3ln3x + 3/x
Therefore, the derivative of f(x) is f′(x) = 3ln3x + 3/x.

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Answer:

Step-by-step explanation:

Complementary Angles add up to 90°

90°-46°=44°

∠m = 44°

Therefore, the solutions of the equation 15x⁴ + x³ - 52x² + 20x + 16 = 0 Reducible to Quadratic Equations are:

x₁ = (±2/√5

x₂ = (±2/√3)

x₃ = 0

x₄ = (±1/20)

To solve 15x⁴ + x³ - 52x² + 20x + 16 = 0 by Equations Reducible to Quadratic Equations, we can use the substitution method. Let's first substitute x² = y and rewrite the equation as:

15y² + y - 52y + 20√y + 16 = 0

Now, let's group the terms:

(15y² - 52y + 16) + (y + 20√y) = 0

Let's solve the first quadratic equation:

15y² - 52y + 16 = 0

We can factor this quadratic equation as:

(3y - 4)(5y - 4) = 0

So, the solutions are:

y₁ = 4/5 and y₂ = 4/3

Now, let's solve the second quadratic equation:

y + 20√y = 0

We can factor out y:

y(1 + 20√y) = 0

So, the solutions are:

y₃ = 0 and y₄ = (-1/20)² = 1/400

Now, let's substitute back y = x²:

x₁ = √(y₁) = ±√(4/5) = ±(2/√5)

x₂ = √(y₂) = ±√(4/3) = ±(2/√3)

x₃ = √(y₃) = 0

x₄ = √(y₄) = ±(1/20)

Therefore, the solutions of the equation 15x⁴ + x³ - 52x² + 20x + 16 = 0 are:

x₁ = (±2/√5)

x₂ = (±2/√3)

x₃ = 0

x₄ = (±1/20)

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It has proved that the equation [tex]e^{(x)} * (dy/dx) = y * e^{(x)} * (dy/dx)[/tex] is linear and the equation [tex](dy/dx) + x^2 * e^{(x)} * y = e^{(x)}[/tex] is not linear.

The equation [tex]e^{(x)} * (dy/dx) = y * e^{(x)} * (dy/dx)[/tex] is linear because it can be written in the standard form of a linear first-order ordinary differential equation, which is:
dy/dx + P(x) * y = Q(x)

In this case, we can divide both sides of the equation by [tex]e^{(x)}[/tex] to obtain:
dy/dx = y * (dy/dx)

Now, we can compare this to the standard form and observe that P(x) = 0 and Q(x) = y * (dy/dx).

Since the equation is in the standard form, it is linear.

On the other hand, the equation [tex](dy/dx) + x^2 * e^{(x)} * y = e^{(x)}[/tex] is not linear.

While it may seem similar to the standard linear form, the presence of the [tex]x^2 * e^{(x)} * y[/tex] term is what makes it non-linear.

In a linear equation, the term involving y should be of the form P(x) * y, where P(x) is a function of x only.

However, in this case, the term [tex]x^2 * e^{(x)} * y[/tex] involves both x and y, making the equation non-linear.

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The line integral of F(8y, 8x) over the circle 32 + y2 = 81 oriented

clockwise is 9.

To compute the line integral of the vector field F(8y, 8x) over the circle

32 + y2 = 81 oriented clockwise, we can use the line integral formula:

∫CF · ds = ∫ab F(r(t)) · r'(t) dt

where C is the curve we are integrating over, F is the vector field, and r(t)

is a parametrization of the curve.

We can parametrize the circle by setting x = 4cos(t) and y = 9sin(t), with t

ranging from 0 to 2π. Then, the curve C becomes:

r(t) = (4cos(t), 9sin(t))

and the unit tangent vector is given by:

T(t) = r'(t)/|r'(t)| = (-4sin(t)/3, 3cos(t)/2)

Note that we divide by the magnitude of r'(t) to get a unit tangent vector.

Then, we can compute the line integral as:

[tex]\int CF ds = \int 0^2 \pi F(r(t)) r'(t) dt[/tex]

= ∫[tex]0^2[/tex]π (8y, 8x) · (-4sin(t)/3, 3cos(t)/2) dt

= ∫[tex]0^2[/tex]π (-32sin(t)cos(t)/3 + 36cos(t)sin(t)/2) dt

= (-16/3)∫[tex]0^2[/tex]π sin(2t) dt + (18)∫[tex]0^2[/tex]π cos(t)sin(t) dt

= (-16/3)[cos(2t)][tex]0^2[/tex]π + (18)[(-1/2)cos2(t)][tex]0^2[/tex]π

= (-16/3)(1-1) + (18)(0+1/2)

= 9

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The answer of the given question based on the commutative property is ,  choice B: c + (a + b).

The commutative property is a fundamental property of some mathematical operations, which states that the order of the operands (inputs) can be changed without affecting the result. In other words, the commutative property means that the operation is independent of the order in which the operands are presented.

The commutative property of addition states that the order of the addends can be changed without changing the sum. In other words, a + b = b + a.

Using this property, we can rearrange the terms in the equation (a + b) + c = A to get:

c + (a + b) = A

This is the same as choice B: c + (a + b). Therefore, the choice that illustrates the commutative property for (a + b) + c is B.

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a) The probability of having at least one major fault in the next 2 km stretch on the highway is approximately 0.959.

b) The distribution of X, the distance between two successive major faults on the highway, is X ~ Exponential (mean=1/1.6). Therefore, the correct option is B.

a) To find the probability of having at least one major fault in the next 2 km stretch on the highway, we first need to find the probability of having zero major faults in that stretch. The number of major faults follows a Poisson distribution with mean λ = 1.6 for every 1 km.

Since we are looking at a 2 km stretch, the new mean is λ' = 2 * 1.6 = 3.2.

Using the Poisson probability mass function (PMF) formula:

P(X = k) = (e^(-λ) * λ^k) / k!

For zero major faults in the 2 km stretch (k = 0):

P(X = 0) = (e^(-3.2) * 3.2^0) / 0! = e^(-3.2)

Now, we want the probability of having at least one major fault, which is the complement of having zero faults:

P(X >= 1) = 1 - P(X = 0) = 1 - e^(-3.2)

Calculating this value:

P(X >= 1) ≈ 1 - 0.040762 = 0.959

So, the probability is approximately 0.959.

b) The distribution of X, the distance between two successive major faults on the highway, is described by an exponential distribution. In this case, the mean distance between major faults is the inverse of the rate (mean number of faults per km).

The rate is λ = 1.6 faults per km, so the mean distance between faults is 1/1.6 km.

Therefore, the correct answer is: B. X ~ Exponential(mean=1/1.6)

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Answer:

8x

Step-by-step explanation:

f(x) + g(x) = (3x-2) + (5x+2)

f(x) + g(x) = 3x + 5x - 2 + 2

f(x) + g(x) = 8x

1. The dimensions of the printed area are approximately:

4.22 in (smaller value)

6.89 in (larger value).

2. An approximation for t, we can use it to find the point on the graph of f(x) that is closest to (x, y):

t = approximate solution from Newton's method

x-coordinate: t

y-coordinate: [tex]t^2 (21).[/tex]

Let the width of the printed area be x and the length of the printed area be y.

Then the total area of the page, including margins, is:

A = (x + 2)(y + 2)

The area of the printed portion is:

xy = 29

We want to minimize the total area, subject to the constraint that the printed area has an area of 29 square inches.

Using the constraint, we can solve for y in terms of x:

y = 29/x

Substituting this into the equation for A, we get:

A = (x + 2)(29/x + 2)

Expanding this expression, we get:

A = 29 + 2x + 58/x + 4

A = 2x + 58/x + 33.

To find the minimum value of A, we take the derivative with respect to x and set it equal to zero:

[tex]dA/dx = 2 - 58/x^2 = 0[/tex]

Solving for x, we get:

[tex]x = \sqrt{(58)}[/tex]

Substituting this back into the equation for y, we get:

[tex]y = 29/\sqrt{(58) }[/tex]

Therefore, the dimensions of the printed area are approximately:

4.22 in (smaller value)

6.89 in (larger value)

To find the point on the graph of the function [tex]f(x) = x^2 (21)[/tex]that is closest to the point (x, y), we can use the distance formula:

[tex]d = \sqrt{((x - t)^2 + (y - f(t))^2) }[/tex]

where t is the value of x that corresponds to the closest point on the graph, and[tex]f(t) = t^2 (21)[/tex]

We want to minimize d, so we take the derivative of d with respect to t and set it equal to zero:

[tex]dd/dt = (x - t) - 2t(21)(y - f(t)) = 0[/tex]

Expanding f(t), we get:

f(t) = 21t^2

Substituting this into the equation for dd/dt, we get:

[tex]dd/dt = (x - t) - 42ty + 42t^3 = 0[/tex]

Solving for t is difficult, but we can use an iterative numerical method, such as Newton's method, to approximate the solution.

We can start with an initial guess, [tex]t_0[/tex] , and use the iteration:

[tex]t_{n+1} = t_n - dd/dt(t_n) / d^2d/dt^2(t_n)[/tex]

where [tex]dd/dt(t_n)[/tex]  is the value of dd/dt at [tex]t_n[/tex], and [tex]d^2d/dt^2(t_n)[/tex] is the second derivative of d with respect to t evaluated at[tex]t_n.[/tex]

We can continue this iteration until the value of [tex]t_n[/tex] stops changing or until we reach a desired level of accuracy.

Once we have an approximation for t, we can use it to find the point on the graph of f(x) that is closest to (x, y):

t = approximate solution from Newton's method.

x-coordinate: t

y-coordinate: [tex]t^2 (21).[/tex]

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It will take approximately 18.61 years for the population to triple.

To solve the initial value problem, we first need to find the value of the constant c. We know that x(0) = 4, so we can substitute this into the equation for x(t) and solve for c:

x(0) = 68/c + 1
4 = 68/c + 1
3 = 68/c
c = 68/3

Now that we know the value of c, we can use the equation for x(t) to find the population after 6 years:

x(6) = 68/(68/3)e^(-0.04×6) + 1
x(6) = 68/(68/3)e^(-0.24) + 1
x(6) = 68/(68/3)(0.789) + 1
x(6) = 3(0.789) + 1
x(6) = 2.367 + 1
x(6) = 3.367

Therefore, the population after 6 years is approximately 3.367.

To find out how long it will take for the population to triple, we need to solve the equation x(t) = 3x(0). Substituting x(0) = 4 and c = 68/3, we get:

3x(0) = 68/(ce^(-0.04t)) + 1
12 = 68/(68/3e^(-0.04t)) + 1
11 = 68/(68/3e^(-0.04t))
11 = 3e^(-0.04t)
ln(11/3) = -0.04t
t = ln(11/3)/(-0.04)

Using a calculator, we find that t is approximately 18.61 years. Therefore, it will take approximately 18.61 years for the population to triple.

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The probability of selecting apple, a peach and a pear in that order is [tex]4.85%[/tex]%.

In this basket, we have 4 apples, 4 peaches, and 4 pears and will select three pieces of fruit without replacement.

The probability of selecting an apple first is 4/12.

The probability of selecting a peach second is 4/11.

The probability of selecting a pear third is 4/10.

P(Apple, Peach, & Pear) = (4/12) * (4/11) * (4/10)

P(Apple, Peach, & Pear) = 64/1320

P(Apple, Peach, & Pear) = 8/165

P(Apple, Peach, & Pear) = 0.04848484848

P(Apple, Peach, & Pear) = 4.85%

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The coordinates of the centroid are:

([tex]\bar x , \bar y[/tex]) = ((3/8)π, (3/8)π)

To find the center of mass of the lamina, we need to find the coordinates of the centroid. Since the density at any point on the lamina is proportional to the square of its distance from the origin, we can use polar coordinates to simplify the calculation.

Let (r, θ) be the polar coordinates of a point on the lamina. Then the density at that point is proportional to [tex]r^2[/tex]. The mass of the lamina can be calculated by integrating the density over the lamina:

m = ∫∫[tex]r^2[/tex] dA

where dA is the differential area element in polar coordinates:

dA = r dr dθ

The limits of integration are 0 to 1 for r and 0 to π/2 for θ, since the lamina is a unit square in the first quadrant.

m = ∫[tex]0^{(\pi /2)[/tex] ∫[tex]0^1 r^4[/tex] dr dθ

= (1/5) ∫[tex]0^{(\pi /2)[/tex] 1 dθ

= π/10

The x-coordinate of the centroid is given by:

[tex]\bar x[/tex] = (1/m) ∫∫x[tex]r^2[/tex] dA

where x is the x-coordinate of a point on the lamina.

Using the fact that x = r cos(θ), we can write:

[tex]\bar x[/tex] = (1/m) ∫∫[tex]r^3[/tex]cos(θ) dASubstituting in the expression for dA in polar coordinates, we have:

[tex]\bar x[/tex] = ((1/m) ∫[tex]0^{(\pi /2)[/tex] ∫[tex]0^1 r^5[/tex]  cos(θ) dr dθ

Evaluating the integral, we get:

[tex]\bar x[/tex] = (3/8)π

Similarly, the y-coordinate of the centroid is given by:

[tex]\bar y[/tex]= (1/m) ∫∫y[tex]r^2[/tex]dA

where y is the y-coordinate of a point on the lamina.

Using the fact that y = r sin(θ), we can write:

[tex]\bar y[/tex]= (1/m) ∫∫[tex]r^3[/tex] sin(θ) dA

Substituting in the expression for dA in polar coordinates, we have:

[tex]\bar y[/tex]= (1/m) ∫[tex]0^{(\pi /2)[/tex] ∫[tex]0^1 r^5[/tex] sin(θ) dr dθ

Evaluating the integral, we get:

[tex]\bar y[/tex] = (3/8)π

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Question

A lamina takes the shape of a unit square (side length of 1) in the first quadrant with one corner at the origin. The density at any point on the lamina is proportional to the square of its distance from the origin: Find the center of mass of the lamina_

It is true that when conducting a t-test for two independent samples with a two-tailed example and α = 0.05, you divide the α by 2 (0.05/2 = 0.025) to find the critical t-scores that form the boundaries of the critical region. This is because you are considering both tails of the distribution.

True. When conducting a t test for two independent samples with a two-tailed example, we need to find the critical t-scores that form the boundaries of the critical region for α = 0.05. To do this, we divide the α value of 0.05 by 2 to get 0.0250, and then use this value to find the t-scores using a t-distribution table or calculator. This is necessary because we are looking at the possibility of a significant difference in either direction, hence the two-tailed example.

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Answer:

43rd term

Step-by-step explanation:

Given arithmetic sequence:

We can use the formula for the nth term of an arithmetic sequence to find which term the number 203 corresponds to in the given arithmetic sequence.

The formula for the nth term of an arithmetic sequence is:

[tex]\boxed{a_n = a_1 + (n-1)d}[/tex]

where:

For the given sequence, we know that the first term is a₁ = -7.

The common difference is d = 5, since each term is 5 more than the previous term.

Substitute these values into the formula to create an equation for the nth term:

[tex]\begin{aligned} \implies a_n &= -7 + (n-1)5\\&=-7+5n-5\\&=5n-12 \end{aligned}[/tex]

To find the position of number 203 in the sequence, substitute aₙ = 203 into the equation for the nth term and solve for n:

[tex]\begin{aligned} a_n &= 203\\ \implies5n-12&=203\\5n-12+12&=203+12\\5n&=215\\\dfrac{5n}{5}&=\dfrac{215}{5}\\n&=43 \end{aligned}[/tex]

Therefore, the number 203 corresponds to the 43rd term in the given arithmetic sequence.

The economic impact of fishing for nearly all great lakes states varies, but according to a report by the U.S. Fish and Wildlife Service, it falls within the range of $1 to $8 billion (in millions of dollars).

This impact includes the economic contributions of recreational fishing, commercial fishing, and related industries such as tourism and boat manufacturing. The exact amount varies from state to state and from year to year depending on factors such as weather, fish populations, and fishing regulations.

The Great Lakes region of the United States is home to some of the largest freshwater bodies in the world and boasts a rich variety of fish species. Fishing is an important economic activity in the region, contributing billions of dollars to the local and national economy. The economic impact of fishing in the Great Lakes region includes not only the direct revenue generated by commercial and recreational fishing, but also the indirect and induced effects of fishing-related industries such as tourism and boat manufacturing.

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1 1/2 feet is equal to 18 inches after the conversion of units of measurement. The correct answer is B) .

One foot is equal to 12 inches. Therefore, to convert 18 inches to feet, we need to divide 18 by 12:

18 inches ÷ 12 inches/foot = 1.5 feet

Option A) 1 1/2 yards is not equal to 18 inches. One yard is equal to 36 inches, so 1 1/2 yards is equal to 54 inches, which is more than 18 inches.

Option C) 2 yards is equal to 72 inches, which is more than 18 inches.

Option D) 2 feet is equal to 24 inches, which is less than 18 inches.

Therefore, the correct answer is B) 1 1/2 feet.

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The mean number of American adults who disapprove of the overall job that President Bush is doing, based on a sample size of 140, is approximately 93.8.

To calculate the mean, we need to first understand what it represents. The mean is a measure of central tendency, which represents the average value of a set of data. In this case, we want to find the average number of American adults who disapprove of President Bush's overall job.

Since we know that 67% of respondents disapproved of President Bush's overall job, we can assume that this percentage also applies to the entire population of American adults.

To find the mean number of American adults who disapprove, we can use the formula:

mean = total / sample size

In this case, the total number of American adults who disapprove can be calculated as:

total = sample size x percentage who disapprove

total = 140 x 0.67

total = 93.8

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Answer: C) option

csc theta is hypotenuse/perpendicular

that is 15/12

Solution for the value cosec θ is,

cosec θ = r / 12

We have,

A right triangle is shown in image,

From the figure,

Apply Pythagoras theorem,

r² = 12² + 9²

r² = 144 + 81

r² = 225

r = 15

Hence, We get;

cosec θ = r / 12

cosec θ = 15 / 12

Therefore, Solution for the value cosec θ is,

cosec θ = r / 12

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Answer:

4

Step-by-step explanation:

80 divided by 4 is 20 then you take 20 and divide it by 5 to get 4 so in conclusion X = 4

Answer:

57,310

Step-by-step explanation:

So you can either round up to 57,310 or round down to 57,300 and 57,309 is one away from 57,310 and 9 away from 57,300 so it’s closer to 57,310 therefore 57,310 is the answer

To round 57,309 to the nearest ten, we look at the ones place digit and round up if the digit is 5 or greater. In this case, since the ones place digit is 9, we round up the tens digit to 1.

To round 57,309 to the nearest ten, we look at the digit in the ones place, which is 9. Since 9 is greater than or equal to 5, we round up the tens digit to the next number. Therefore, 57,309 rounded to the nearest ten is 57,310.

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A. Decimal squares and C. Base ten pieces would be the foremost reasonable manipulative materials for instructing decimal documentation to the hundredths put.

Decimal squares can offer assistance to understudies to visualize the relationship between tenths, hundredths, and thousandths, as each square can be separated into 10 little squares. Understudies can utilize squares to construct and compare decimal numbers to the hundredths put.

Base ten squares can too be utilized to speak to decimals to the hundredths put, with one level speaking to one entirety, one bar speaking to one-tenth, and one unit speaking to one-hundredth. Understudies can construct and compare decimal numbers utilizing the pieces, as well as utilize them to show operations with decimals.

The other manipulative materials recorded (Design pieces, Tangrams, Color Tiles, and Geoboards) are not particularly planned to speak to decimals and would likely not be as viable in instructing decimal documentation to the hundredths put. 

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The total distance traveled by the airplane is 15.3 feet.

The Pythagorean theorem asserts that the square of the length of the hypotenuse in a right triangle equals the sum of the squares of the lengths of the two legs in a right triangle. This theorem is the basis for the distance formula. In the distance formula, the hypotenuse is the distance between the two points, and the two legs are the differences between the x- and y-coordinates of the two points. Any two locations in a two-dimensional coordinate system can have their distance between them calculated using the formula.

Let us suppose the starting point, that is, the point for jack as (0, 0).

Now, according to the given placements the position of the other students are:

Destiny: (0,2)

Barbara: (-9,2)

Now, using the distance formula we have:

The distance between Jack and Destiny is:

√[(0-0)² + (2-0)²] = √(4) = 2

The distance between Destiny and Barbara is:

√[(-9-0)² + (2-2)²] = √(81) = 9

The distance between Barbara and Jack is:

√[(0-(-9))² + (0-2)²] = √(85)

So the total distance traveled by the paper airplane is:

2 + 9 + √(85) ≈ 15.3 feet

Hence, the total distance traveled by the airplane is 15.3 feet.

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Hazel will spend $2.58 on 0.9 pounds of coffee ice cream and 3.4 pounds of cherry ice cream at a price of $0.6 per pound.

To determine how much Hazel will spend, we need to multiply the weight of each ice cream flavor by its price per pound and then add the two products together

Cost of coffee ice cream = 0.9 pounds × $0.6/pound = $0.54

Cost of cherry ice cream = 3.4 pounds × $0.6/pound = $2.04

Total cost = $0.54 + $2.04 = $2.58

Therefore, Hazel will spend $2.58 on 0.9 pounds of coffee ice cream and 3.4 pounds of cherry ice cream at a price of $0.6 per pound.

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--The given question is incomplete, the complete question is given

" If Hazel buys 0. 9 pounds of coffee ice cream and 3. 4 pounds of cherry ice cream, and 0.6 pound per price. how much will she spend?"--

Measures of m∠1 = 67.4°, m∠2 = 104.5°

What are the measures of angle?

When two lines or rays intersect at a single point, an angle is created. The vertex is the term for the shared point. An angle measure in geometry is the length of the angle created by two rays or arms meeting at a common vertex.

Here, we have

The exterior angle 121.8° is the sum of the remote interior angles 17.3° and angle 2. Then ...

angle 2 = 121.8° -17.3° = 104.5° . . . . . . . matches the second choice

Angle 2 is the exterior angle of the top triangle. It, too, is the sum of the remote interior angles:

104.5° = angle 1 + 37.1°

angle 1 = 104.5° -37.1° = 67.4°

Hence, the measures of m∠1 = 67.4°, m∠2 = 104.5°.

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Answer:

60

Step-by-step explanation: