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2 years ago

The area of a rectangle is 1,357 square feet. The length is 59 feet. What is the width?

Mathematics

2 answers:

Marysya12 [62]2 years ago

8 0

Answer:

Step-by-step explanation:

dangina [55]2 years ago

3 0

Answer:

w = 23

Step-by-step explanation:

a = l × w

The area and length are given, so you can substitute them in.

1357 = 59 × w

Solve for w:

w = $\frac{1357}{59}$

w = 23

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An airline wants to evaluate the depth perception of its pilots over the age of fifty. A random sample of n = 14 airline pilots

Y_Kistochka [10]

Answer:

Step-by-step explanation:

Hello!

1) The objective is to evaluate the depth perception of pilots over the age of fifty.

To do so a sample of n=14 airline pilots over the age of fifty was taken, each of them was asked to judge the distance between two markers placed 20 feet apart, the pilot's error in judging the distance was recorded:

2.9, 2.6, 2.9, 2.6, 2.4, 1.3, 2.3, 2.2, 2.5, 2.3, 2.8, 2.5, 2.7, 2.6

Then the variable of interest is X: error in judging the distance between two markers placed 20 feet apart of one pilot. (feet)

Assuming that this variable has a normal distribution, with a standard deviation of σ= 1.4 feet

The hypothesis is that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1, symbolically: μ > 2.1

H₀: μ ≤ 2.1

H₁: μ > 2.1

α: 0.05

Since the variable has a normal distribution and the population standard deviation is known, the statistic to use for this test is the standard normal:

$Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

The sample mean is

X[bar]= ∑X/n= 34.90/14= 2.49

$Z_{H_0}= \frac{2.49-2.1}{\frac{1.4}{\sqrt{14} } } = 1.04$

The p-value for this test is 0.14917

Using the p-value approach, since it is greater than the significance level, the decision is to not reject the null hypothesis.

Then there is no evidence that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1 feet.

2) To construct the 90% confidence interval you have to use the same distribution as before, the formula for the interval under the standard deviation is:

[X[bar] ± $Z_{1-\alpha /2}$ * (δ/√n)]

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[2.49 ± 1.645 * (1.4/√14)]

[1.87; 3.11]

With a 90% confidence, you'd expect that the true mean of the error in-depth perception of the airline's pilots over fifty years old is contained in the interval [1.87; 3.11].

I hope you have a SUPER day!

7 0

2 years ago

Toms math teacher asks 6 is 20 percent of what number tom incorrectly says 120 what’s is the correct number

Setler79 [48]

6=20% * n

6 = .20 * n

divide by .20

6/.2 = n

30 =n

Answer: 30

8 0

2 years ago

If ABC = ADC, which is NOT true by CPCTC ?

sdas [7]

We cannot use CPCTC to show that BE is congruent to DE.

CPCTC refers to the angles and sides of two congruent triangles being congruent to their corresponding pieces. The height of a triangle is not one of the given sides that counts in this.

4 0

2 years ago

A veterinary study of horses looked water sources for horses and the investigators found that horses received water from either

Irina18 [472]

Answer:

d. Water Source Well City Water Stream Expected Count 45.33 45.33 45.33

Step-by-step explanation:

Hello!

The objective of this study is to test whether the horses are equally likely to get water from one of three sources (Well, City or Stream), i.e. that the water sources have the same probability of being used.

The study variable is categorical:

X: Origin of the drinking water of the horses, categorized: Well "W", City "C" and Stream "S"

And one study population:

"Horses"

If the study hypothesis is that the three water sources are equally to be selected, you can symbolize it as P(W)=P(C)=P(S)=1/3

You want to test if the observed distribution follows an expected distribution, the test to use is a Chi-Square Goodness of Fit Test.

The hypothesis is:

H₀: P(W)=P(C)=P(S)=1/3

Under the null hypothesis, the formula to calculate the expected frequencies is:

$e_{i}= n * P_{i}$

Where:

$e_{i}$ is the expected value of the i-category.

$P_{i}$ is the probability for the i-category under the null hypothesis.

n= sample size.

Observed frequencies:

Well: 62

City: 43

Stream: 31

Total: 136

Expected frequencies:

$e_{W}= n * P_{W} = 136*1/3 = 45.33$

$e_{C}= n * P_{C}= 136 * 1/3= 45.33$

$e_{S}= n * P_{S}= 136 * 1/3= 45.33$

Typ, if the expected frequencies are correctly calculated, the sum of them must give the total of the sample (although depending on the type of approximation it can vary in one or two decimals)

45.33 * 3= 135.99 ≅ 136

I hope it helps!

5 0

2 years ago

A human body while sleeping generates about 300 BTUs of body heat per hour. If two people are sleeping in a bedroom that is 15 f

Katarina [22]

Given:
one person can produce:
E = 300 BTUs of body heat per hour
dimensions of the bedroom is:
V = 15 ft × 12 ft × 10 ft

required: is the energy per volume if 2 persons sleeps in 7 hours
let T be the total energy produce per volume

T = (300 BTU per hour per person) ( 7 hours) ( 2 person) / ( 15 ft × 12 ft × 10 ft)

T = 2.3 BTU / ft^3

4 0

2 years ago