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GrogVix [38]
2 years ago
6

Hiro painted his room at a rate of 8 square meters per hour. After 3 hours of painting, he had 28 square meters left to paint. L

et A(t) denote the area to paint A (measured in square meters) as a function of time t (measured in hours).
Mathematics
1 answer:
lesya692 [45]2 years ago
4 0

Answer: A(t)=-8t+52

Step-by-step explanation:

<h3> The missing question is: "What is the Functions formula A(t)=?"</h3><h2 />

The  equation of the line in Slope-Intercept form is:

y=mx+b

Where "m" is the slope and "b" is the y-intercept.

According to the data given in the exercise, you know that:

- A(t) represents the area to paint the Hiros' romm as a function of time.

- The rate he painted the room was 8 square meters per hour.

- The area left to paint after 3 hours was 28 m².

Therefore, based on this, you can idenfity that:

1. The slope of the line is:

m=-8

2. One of the point on the line is:

(3,28)

So you must substitute the slope and the coordinates of that point into y=mx+b and then solve for "b" in order to find its value:

28=-8(3)+b\\\\28+24=b\\\\b=52

Therefore, you can determine that the function A(t) is:

A(t)=-8t+52

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Suppose the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations. Th
madreJ [45]

Answer:

<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.

Step-by-step explanation:

A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.

It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.

The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:

<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.

<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.

The significance level of the test is:

<em>α</em> = 0.05.

The test is defined as:

z=\frac{(\bar x_{1}-\bar x_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}

The decision rule for the test is:

If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.

6 0
2 years ago
A food scientist takes a random sample of 25 cereals in order to test the claim that the mean fiber content is less than 2.5 gra
Jlenok [28]

Answer:

Choice A: there is insufficient evidence to conclude that the mean for the fiber content is less than 2.5 gms.

Step-by-step explanation:

From the graphs given the more appropriate is one sample t- test

One sample t- test is used and the results are

Hypothesis Test                       Fiber

Sample Mean                          2.62

Sample St. Dev                       3.492

Degrees of Freedom              24

t- test statistic                         0.1718

p- value                                    0.5675

the critical region for this test at ∝= 0.05 is t < -1.711

There is not enough evidence to reject the null hypothesis.

Since the calculated t=   0.1718 does not fall in the critical region the null hypothesis is accepted that the mean fiber content is not  less than 2.5 grams.T

6 0
2 years ago
A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer
choli [55]
<h2>Answer:</h2>

The area of the top surface of the washer is:  160.14 square mm.

<h2>Step-by-step explanation:</h2>

The top of the surface is in the shape of a annulus  with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

Area=\pi (R^2-r^2)

where R is the outer radius and r is the inner radius.

Here we have:

R=10\ mm\\\\and\\\\r=7\ mm

Hence, we have:

Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2

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Answer:

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