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andrew-mc [135]

2 years ago

7

Bradley has ordered a steak from the meat market that weighs 1 5/8 pounds. There is a bone in the steak that weighs 7/8 of a pou

nd. How much does the meat weigh?

1 answer:

KonstantinChe [14]2 years ago

6 0

Answer:

3/4

Step-by-step explanation:

the bone and the steak weigh = 1 5/8 pounds

The bone weighs 7/8

The meat weighs = 15/8 - 7/8

Converting 1 5/8 to an improper fraction gives 13/8

13/8 - 7/8 = 6/8 = 3/4

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The standard deviation of the set of values (2, 3, 6, 9, 10) is <span>square root of 12.5 (</span> $\sqrt{12.5}$ ) therefore OPTION C

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nikita wrote the expression 72 + 26. Bettina wrote the expression 4(72+26). Which sentence is true about the value of their expr

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Nikita's value is 4 times less than Bettina's value. I believe this is the answer.

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3 years ago

From where Kiana is standing, she must look up at a 50° angle to see the top of a building. If she backs up 50 feet, she will ne

Morgarella [4.7K]

For this case, what we must do is solve the following system of equations:
tan (50) = h / x
tan (40) = h / (x + 50)
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(x + 50) * tan (40) = h
(x) * tan (50) = h
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x (tan (50) - tan (40)) = 50 * tan (40)
x = 50 * tan (40) / (tan (50) - tan (40))
x = 118.9692621
Substituting:
h = (x) * tan (50)
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Answer:
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5 0

2 years ago

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In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago

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scoundrel [369]

Answer:

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Step-by-step explanation:

8 0

3 years ago