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2 years ago

Mei is trying to list as many prime numbers as she can for a challenge problem for her math class. She finds

Mathematics

1 answer:

Svet_ta [14]2 years ago

6 0

Hello,

Since the degree of a polynomial is the highest exponent present, the degree of Mei’s polynomial should be 2.

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Find the volume of the composite solid figure<br>(picture provided of the shape)

LiRa [457]

Answer:

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2 years ago

Step by step, to find solution to -4(6n-4)+3n=

Dimas [21]

Answer:

-21n + 16

Step-by-step explanation:

Start with distributive property

-24n + 16 + 3n

Combine like terms

-21n + 16

There is a part of the equation missing so I can't solve completely

7 0

2 years ago

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Any Ideas on How to solve this Question? And if possible the Solution itself.

sweet [91]

The lower quartile is 60. Quartiles are the numbers in the middle of the median and lowest/highest number.

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3 years ago

Slope is known as the run over the rise. True or False

pashok25 [27]

Answer:

false. slope is calculated by rise over run

8 0

2 years ago

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The travel-to-work time for residents in Ventura County is unknown. Assume the population variance is 39. How large should a sam

katen-ka-za [31]

Answer:

A sample size of 128 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.93}{2} = 0.035$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.035 = 0.965$ , so $z = 1.81$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population(square root of the variance) and n is the size of the sample.

How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

We need a sample size of n, which is found when $M = 1$ . We have that $\sigma = \sqrt{39}$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.81*\frac{\sqrt{39}}{\sqrt{n}}$

$\sqrt{n} = 1.81\sqrt{39}$

$(\sqrt{n})^2 = (1.81\sqrt{39})^2$

$n = 127.8$

Rounding up

A sample size of 128 is needed.

7 0

2 years ago