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3 years ago

7

a rectangle has a perimeter of 3o inches and a width of 6 inches what are the dimensions of a similar rectangle with a perimeter

of 20 inches?

1 answer:

xz_007 [3.2K]3 years ago

3 0

Answer:

the width would be 4 inches and the length would be 6 inches

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A blimp, suspended in the air at a height of 600 feet, lies directly over a line from a sports stadium to a planetarium. If an a

Molodets [167]

Step-by-step explanation:

so, this sounds like the blimp is located between the stadium and the planetarium.

we have a triangle :

the ground distance between the stadium and the planetarium is the baseline.

and the 2 lines of sight from the blimp on one side to the stadium and on the other side to the planetarium are the 2 legs.

we know the height of this triangle is 600 ft.

the angle of depression down to the stadium is 37°. which makes the inner triangle angle at the ground point at the stadium also 37°.

and the angle of depression down to the planetarium is 29°. which makes the inner triangle angle at the ground point at the planetarium also 29°.

and because the sum of all angles in a triangle is always 180°, we know the angle at the blimp is

180 - 37 - 29 = 114°

in order to solve this triangle, we need to split it into 2 right-angled triangles by using the height of the main triangle as delimiter.

we get a stadium side and a planetarium side triangle.

the baselines (Hypotenuses) of the 2 triangles are the corresponding lines of sight from the blimp.

the height of the large triangle is also a height and a leg in each small triangle.

and the stadium side part of the large baseline (between ground point and intersection with the height) is the second leg for the stadium side triangle.

and correspondingly, the planetarium side part of the large baseline (between ground point and intersection with the height) is the second leg for the planetarium side triangle.

the inner blimp angle of the stadium side triangle is

180 - 37 - 90 = 53°

and the inner blimp angle of the planetarium side triangle is

180 - 29 - 90 = 61°

now we can use the law of sine to get the lengths of the 2 parts of the baseline of the large triangle.

and when we add these 2 numbers we get the distance between stadium and planetarium.

law of sine is

a/sin(A) = b/sin(B) = c/sin(C)

with the sides being opposite of the associated angles.

for the stadium side triangle we get

part1/sin(53) = 600/sin(37)

part1 = 600×sin(53)/sin(37) = 796.226893... ft

for the planetarium side triangle we get

part2/sin(61) = 600/sin(29)

part2 = 600×sin(61)/sin(29) = 1,082.428653... ft

the distance between the stadium and the planetarium is

part1 + part2 = 1,878.655546... ft

4 0

2 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

Bryce has 220 feet of fencing that will enclosure a rectangular corral. One side of the corral will be 48 feet long. What will b

Gala2k [10]

Answer:

10560 square feet

Step-by-step explanation:

If 2 sides are 48 feet, then we are left with

$220-2*48=124$ feet for the remaining side pieces.

$\frac{124}{2} =62$ feet.

So our dimensions must be 48 x 62.

This gives us an area of

$220*48=10560ft^{2}$

6 0

3 years ago

Twice a number is equal to 35 more than seven times the number. Find the number

amm1812

number - n

Twice a number - 2n

Seven times the number - 7n

2n = 7n +35

7n-2n = -35

5n = -35

n = - 7

Check:

2(-7) = 7(-7) +35

-14 = -49 + 35

-14 = -14 True

The number is (- 7).

8 0

4 years ago

What is the volume of the cone

Umnica [9.8K]

Answer:

37.68

Step-by-step explanation:

use the formula pi(3.14)r^2 h/3

3 0

3 years ago