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Andru [333]
4 years ago
13

Second time...40 pts! Find the surface area of the model below (image included).

Mathematics
1 answer:
charle [14.2K]4 years ago
5 0
5•8=40•2=80+(8•8)=144
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What is lcm of 54 and 64
ankoles [38]
The lcm is 8

7×8=56

6×6=64


7 0
3 years ago
Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa
german
\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see 0.145\times7500\approx10.86 scores above 600.
5 0
3 years ago
Please answer this I need it for a test please
salantis [7]

Answer:

20

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for
Setler [38]

Answer:

7) d)

standard error of the mean of one sample of 'n' observation = 0.20

8) a)

The margin of Error = 0.392

9) d

The 95% of confidence intervals are (8.61 , 9.39)

Step-by-step explanation:

7)

solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

The standard error of the mean of one sample of 'n' observation is

Standard error (SE)

                               = \frac{S.D}{\sqrt{n} }  

                               = σ / √n

                               = \frac{1.8}{\sqrt{81} } =0.2

standard error of the mean of one sample of 'n' observation = 0.20

8)

Solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

Given the probability is 0.95

The z- score = 1.96 at 0.05 level of significance.

The margin of Error   =  \frac{z_{0.95} S.D}{\sqrt{n} }

                                   = \frac{1.96 (S.D)}{\sqrt{n} }

                                   = \frac{1.96 (1.8)}{\sqrt{81} }

                                   = 0.392

The margin of Error = 0.392

9)

Solution:-

<u>The 95% of confidence intervals are </u>

<u></u>(x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96\frac{S.D}{\sqrt{n} } )<u></u>

<u></u>(9 - 1.96\frac{1.8}{\sqrt{81} } , 9+ 1.96\frac{1.8}{\sqrt{81} } )<u></u>

(9 - 0.392 , (9 + 0.392)

(8.609 , 9.392)

<u>The 95% of confidence intervals are </u>(8.61 , 9.39)

 

4 0
3 years ago
These two lines represent a system of equations.<br> What is the solution to that system?<br> Help ?
Mars2501 [29]

Answer:

yhuvguvcxdcfvgbhnjmkj

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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