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3 years ago

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A person wants to increase a (5 in. by 7 in.) photo to an (8 in. by 10 in.) but since the aspect ratios are not the same some of

the picture will get chopped off. What percentage of the picture can be used in the (8 in. x 10 in.) frame?

1 answer:

Llana [10]3 years ago

3 0

Answer:

89.3

Step-by-step explanation:

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Which fraction belongs in the box to make the expression true? 1/4 < ___< 7/8. A. 1/5. B. 8/9. C. 2/10. D. 10/20

Yuri [45]

The answer is D.

In the expression, the fraction must be larger than 1/4, yet smaller than 7/8.

It can't be A because 1/5 is smaller than 1/4. See below.

1/5 = 4/20

1/4 = 5/20

It also can't be B because 8/9 is larger than 7/8. See below.

8/9 = 64/72

7/8 = 63/72

Lastly, the answer cannot be C because 2/10 is smaller than 1/4.

1/4 = 5/20

2/10 = 4/20

Answer D works because:

10/20 = 2/4 > 1/4

10/20 = 80/160 < 140/160

7 0

3 years ago

Read 2 more answers

vitfil [10]

Answer:

45-(-15)

430-(-180)

11,560-(-185)

Step-by-step explanation:

just add the two numbers together for diffrence.

6 0

2 years ago

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musickatia [10]

Answer:

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Step-by-step explanation:

3 0

3 years ago

A construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. If the buildi

Vlad1618 [11]

Let x represent the height of the model.

We have been given that a construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. We are asked to find the height of the model, if the building is expected to be 200 feet tall.

We will use proportions to solve our given problem as:

$\frac{\text{Model height}}{\text{Actual height}}=\frac{\text{Model length}}{\text{Actual length}}$

Upon substituting our given values, we will get:

$\frac{x}{200\text{ ft}}=\frac{3\text{ in}}{\text{32 ft}}$

$\frac{x}{200\text{ ft}}\times 200\text{ ft}=\frac{3\text{ in}}{\text{32 ft}}\times 200\text{ ft}$

$x=3\text{ in}\times 6.25$

$x=18.75\text{ in}$

Therefore, the model will be 18.75 inches tall.

7 0

2 years ago

Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace

joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean $\mu$ and default $\sigma$ , then the mean take seriously given as the straight line with a z score given by the confidence interval

$\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\$

Using formula:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

$P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)$

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

$=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z$

the required probability: $P(z>0.8349)=1-P(z$

4 0

2 years ago