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2 years ago

Is it wrong to get banned from a school discord

Mathematics

2 answers:

Cerrena [4.2K]2 years ago

7 0

Yes but If you did something not that bad then no

Nataly_w [17]2 years ago

6 0

Answer:

absolutely

Step-by-step explanation:

nice going if u did.. im curious what for though

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Why is 1 3 + 1 3 NOT equal to 2 6 ?

katrin [286]

It is a trick question 13+13 DOES equal 26

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3 years ago

recent survey found that 71% of all adults over 50 wear glasses for driving. In a random sample of 30 adults over 50, what is

valentina_108 [34]

Answer:

Mean = 21.3

Standard Deviation = 2.48

Step-by-step explanation:

We are given the following in the question:

p = 71% = 0.71

Sample size, n = 30

We have to find the mean and the standard deviation of those that wear glasses.

$\text{Mean} = np\\=30(0.71)\\=21.3$

Thus, the mean of those who wear glasses is 21.3

$\text{Standard deviation} = \sqrt{npq}\\=\sqrt{np(1-p)}\\=\sqrt{30\times 0.71\times (1-0.71)}\\=2.48$

Thus, the standard deviation of those who wear glasses is 2.48

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3 years ago

Please help!! will give brainliest

Romashka [77]

Answer A is the only one that seems to make sense to me id choose A

8 0

3 years ago

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A Confidence interval is desired for the true average stray-load loss mu (watts) for a certain type of induction motor when the

Vaselesa [24]

Answer:

A sample size of 35 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:

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

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 $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the width W as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large must the sample size be if the width of the 95% interval for mu is to be 1.0:

We need to find n for which W = 1.

We have that $\sigma^{2} = 9$ , then $\sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3$ . So

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

$1 = 1.96*\frac{3}{\sqrt{n}}$

$\sqrt{n} = 1.96*3$

$(\sqrt{n})^2 = (1.96*3)^{2}$

$n = 34.57$

Rounding up

A sample size of 35 is needed.

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3 years ago

NEED HELP!!! BRANLIEST FOR CORRECT ANSWER FIRST!!!

amid [387]

root 30 ..................

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3 years ago

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