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3 years ago

Please help WILL MARK BRAINLIEST !!!!

Mathematics

2 answers:

Marizza181 [45]3 years ago

7 0

Hello hello hello hello

adell [148]3 years ago

3 0

Answer:

100 ft squared

Step-by-step explanation:

5 times 5 for each square

25 times 4

100 ft squared

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"A curtain manufacturer uses a 90-yard roll of material to make 77 identical curtains. There was 7.75 yards of material left ove

goblinko [34]

Answer:

1.06818181 there you go

7 0

3 years ago

Read 2 more answers

The lowest elevation in Louisiana is 8 feet below sea level the highest version is 535 feet above sea level what is the differen

ELEN [110]

8(highest)+535(lowest)=543

an easy way to get the answer is to just add the two numbers together because your just trying to find how far it is between -535 and a positive 8.

5 0

3 years ago

The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied y

ioda

Answer:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

$s= \sqrt{856849.7}=925.662$

Step-by-step explanation:

For this case we can create the following table

Interval Frequency (f) Midpoint(xi) xi *f xi^2* f

0-499 9 249.5 2245.5 560252.3

500-999 13 749.5 9743.5 7302753

1000-1499 33 1249.5 41233.5 51521258.25

1500-1999 115 1749.5 201193.5 361986278.8

2000-2499 125 2249.5 281187.5 632531281.3

2500-2999 81 2749.5 222709.5 612339770.3

3000-3499 47 3249.5 152726.5 496284761.8

3500-3999 45 3749.5 168727.5 632643761.3

4000-4499 22 4249.5 93489 397281505.5

4500-4999 10 4749.5 47495 225577502.5

Total 500 1220750 3408029125

$\sum f_i = 500 , \sum x_i f_i = 1220750, \sum x^2_i f_i = 3408029125$

For this case we can calculate the sample mean with this formula:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

And for the sample variance we can use the following formula:

$s^2= \frac{\sum x^2_i f_i - \frac{(\sum x_i f_i)^2}{n}}{n-1}$

And if we replace we got:

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

And the deviation is just the square root of the sample variance and for this case is:

$s= \sqrt{856849.7}=925.662$

4 0

3 years ago

The desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5

LekaFEV [45]

Given that the desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. suppose that the percentage of sio2 in a sample is normally distributed with σ = 0.32 and that $\bar{x}=5.24$ .


To investigate whether this indicate conclusively that the true average percentage differs from 5.5.

Part A:

From the question, it is claimed that the desired average percentage of sio2 in a certain type of aluminous cement is 5.5 and we want to test whether the information from the random sample indicate conclusively that the true average percentage differs from 5.5.

Therefore, the null hypothesis and the alternative hypothesis is given by:

$H_0:\mu=5.5 \\ \\ H_a:\mu\neq5.5$

Part B:

The test statistics is given by:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\ \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08} \\ \\ =-3.125$

Part C:

The p-value is given by

$P(z\ \textless \ -3.125)=1-P(z$

Part D:

Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."

Part E:

If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H0? (Round your answer to four decimal places.)

The probability of detecting the departure from $H_0$ is given by

$1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\ \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\ \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)$

$=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\ \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\ \\ =1-\phi(1.326)+\phi(-3.826) \\ \\ =1-0.90758+0.00007 \\ \\ =0.0925$

Part F:

What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.)

The value of n is required to satisfy α = 0.01 and β(5.6) = 0.01 is given by

$n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\ \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\ \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\ \\ =(15.6864)^2=247$