A. The graph is translated 2units down.

A contractor has been hired to install flooring for a bandstand in the shape of a circle that is 21 feet across. How many square

SOVA2 [1]

Answer:

A. 346.19 square feet.

Step-by-step explanation:

We have been given that a contractor has been hired to install flooring for a bandstand in the shape of a circle that is 21 feet across.

To find the square feet of flooring required, we will find area of circle.

Since the length across the circle is 21 feet, so radius of circle will be half of 21 feet.

$\text{Radius}=\frac{\text{21 feet}}{2}$

$\text{Radius}=\text{10.5 feet}$

$\text{Area of circle}=\pi r^2$

$\text{Area of circle}=\pi (\text{10.5 feet})^2$

$\text{Area of circle}=3.14*110.25\text{ feet}^2$

$\text{Area of circle}=346.185\text{ feet}^2$

$\text{Area of circle}\approx 346.19\text{ feet}^2$

Therefore, 346.19 square feet of flooring will be required.

3 0

3 years ago

Read 2 more answers

If there were 1000 dominos for the price of $119.99 how much is each domino?

Novay_Z [31]

Answer:

its either $8.34 or $8

Step-by-step explanation:

8 0

2 years ago

Baby Amelia's parents measure her height every month.

jonny [76]

The statement H(30) = H(25) + 5 means that, "when Amelia was 30 months old, she was 5 centimeters taller than when she was 25 months old"

<h3><u>Solution:</u></h3>

Given that,

Baby Amelia's parents measure her height every month

It is given that H(t) models Amelia's height (in centimeters) when she was "t" months old

Given statement is H(30) = H(25) + 5

So, we can say that H(30) = Amelia's height when she was 30 months old

H(25) = Amelia's height when she was 25 months old

So H(30) = H(25) + 5 means that,

Height of Amelia when she was 30 months old is 5 centimeter more than Amelia height when she was 25 years old

Or we can say that,

When Amelia was 30 months old, she was 5 centimeters taller than when she was 25 months old

7 0

3 years ago

39.76 divided by 2.8

denpristay [2]

Answer:

14.2

Step-by-step explanation:

6 0

2 years ago

An auto company claims that the fuel efficiency of its sedan has been substantially improved. A consumer advocate organization w

zzz [600]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

$df=n-1=12-1=11$

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$ , we see that $p_v < \alpha$ , so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation :

x=test value 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.685$

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(11)}>2.021) =0.0342$

If we compare the the p value with the significance level provided $\alpha=0.1$ , we see that $p_v < \alpha$ , so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

4 0

3 years ago