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3 years ago

What is the width of a rectangle with length 5.5 cm and area 220 cm???

Mathematics

1 answer:

Aleks04 [339]3 years ago

7 0

40 cm. Just divide the area by the length. Hope this helps!

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Help me with this pls

ella [17]

Answer:

7:3

Step-by-step explanation:

Divide both sides by 4

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3 years ago

100 pts for a few questions Ill give brainliest! (if you answer 5 ?s you get the other 50)

spayn [35]

Answer: ik nothing of basketball sorry

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3 years ago

Read 2 more answers

A bag contains five red marbles, fifteen black marbles, and ten white marbles. You pick one without looking. What is the probabi

aliya0001 [1]

Answer:

Step-by-step explanation:

5 red

15 black

10 white

altogether the is 30 marbles so that means it is 'out of' 30

and because the is 5 red it would be 5/30

and because the is 10 white it would be 10/30

that should be the answer but if you have to work the total all together then it would be 15/30

hope this helps :)

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3 years ago

What is the y-intercept of the function f(x)=4 - 5x?<br> -5<br> -4<br> 4<br> 5

pantera1 [17]

Answer:

b = 4

Step-by-step explanation:

Use the slope-intercept form y = m x + b to find the y-intercept b .

8 0

3 years ago

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As a follow-up to a report on gas consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage f

kondor19780726 [428]

Answer:

The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).

Step-by-step explanation:

Our sample size is 96.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 96-1 = 95$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 95 and 0.025 in the t-distribution table, we have $T = 1.9855$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{5.6}{\sqrt{96}} = 0.57$

Now, we multiply T and s

$M = T*s = 1.9855*0.57 = 1.31$

For the lower end of the interval, we subtract the mean by M. So $19.6 - 1.31 = 18.29$

For the upper end of the interval, we add the mean to M. So $19.6 + 1.31 = 20.91$

The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).

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4 years ago