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Karo-lina-s [1.5K]
3 years ago
10

find the area of a rectangle that has a length of 8.3 inches. and a width that is 0.6 times the length. show work please

Mathematics
2 answers:
grigory [225]3 years ago
6 0
Find the width first...8.3×0.6.....then find the area....length × width= 8.3×4.9=40.67....that's what I got
ValentinkaMS [17]3 years ago
6 0
Area=length×width
8.3
×0.6
=====
6×3=18 put down 8 and carry the
6×8=48 + the 1 =498
498...now count how many numbers are behind the decimals...2 = 4.98 4.98 × 8.3 = 41.334in (2)
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(2×4×4)+(3×2×4)

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When Vlad moved to his new home a few years ago, there was a young oak tree in his backyard. He measured it once a year and foun
Anestetic [448]

Answer:

The tree was 175 centimeters tall when Vlad moved into the house.

7 years passed from the time Vlad moved in until the tree was 357 centimeters tall.

Step-by-step explanation:

The height of the tree, in centimeters, in t years after Vlad moved into the house is given by an equation in the following format:

H(t) = H(0) + at

In which H(0) is the height of the tree when Vlad moved into the house and a is the yearly increase.

He measured it once a year and found that it grew by 26 centimeters each year.

This means that a = 26

So

H(t) = H(0) + 26t

4.5 years after he moved into the house, the tree was 292 centimeters tall. How tall was the tree when Vlad moved into the house?

This means that when t = 4.5, H(t) = 292. We use this to find H(0).

H(t) = H(0) + 26t

292 = H(0) + 26*4.5

H(0) = 292 - 26*4.5

H(0) = 175

The tree was 175 centimeters tall when Vlad moved into the house.

How many years passed from the time Vlad moved in until the tree was 357 centimeters tall?

This is t for which H(t) = 357. So

H(t) = H(0) + 26t

H(t) = 175 + 26t

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2 years ago
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3 years ago
A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const
Ludmilka [50]

Answer:

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation  

s=0.68 represent the sample standard deviation  

\bar x =98.90 represent the sample mean  

n=98 the sample size  

Confidence=90% or 0.90  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

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The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:  

\chi^2_{\alpha/2}=120.990  

\chi^2_{1- \alpha/2}=75.282  

And replacing into the formula for the interval we got:  

\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}  

0.371 \leq \sigma^2 \leq 0.596  

Now we just take square root on both sides of the interval and we got:  

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

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