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2 years ago

A number is added to 9. The result is then multiplied by 4 to give a new result of

Mathematics

2 answers:

Ivanshal [37]2 years ago

7 0

Answer:

the number is 21

Step-by-step explanation:

(9+x)(4)=120

9+x=30

x=21

stich3 [128]2 years ago

5 0

Answer:

A, 21

Step-by-step explanation:

You would first need to divide 4 by 120 to get an idea on what a number plus 9 would get you.

If you divide 4 by 120, it'll be 30. So thats what the answer would be when the number is added to 9.

Now that we have 30, we can subtract 9 by it which will give you 21.

When you double check, 9+21=30, 30 x 4=120.

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The heights of women aged 20 to 29 follow approximately the N(64, 2.82) distribution. Men the same age have heights distributed

ivann1987 [24]

Answer:96.97

Step-by-step explanation:

Given

mean height of women $\mu _w=64$

Standard deviation of height $\sigma _w=2.82$

mean height of men $\mu _=69.3$

Standard deviation of height $\sigma _m=2.71$

P $\left ( X$

P $\left ( \frac{x-\mu _w}{\sigma _w}\right )$

P $\left ( Z$

i.e. 0.9697=96.97 %

4 0

2 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.