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3 years ago

I will give brainiest to whoever answers correctly !!

Mathematics

1 answer:

Semmy [17]3 years ago

4 0

$157.50 because if you use the simple interest formula (i=p*r*t), you get $157.50. Hope this helped with your homework!!

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How many times greater is 130,000,000 than 64,800,000?

insens350 [35]

$\frac{130,000,000}{64,800,000} = \frac{1,300}{648} =\boxed {2.00617283951}$

4 0

3 years ago

Does the table represent a function ? If so, state the domain and range. If not, state why. (Picture)

andrezito [222]

Yes

For f(x) to represent a function each value of x from the domain must only map to one unique value in the range, called one-to-one

This is the case here as each value of x maps to one unique value in the range

domain x ∈ {- 5, - 4, 0, 3 }

range y ∈ {0, 2, 10, 16 }

7 0

3 years ago

Will mark Brainlest Help please<br>If a:b =4:5 then find the duplicate value of a:b<br>

Nostrana [21]

Answer:

it must be 5:4..I'm not sure.

4 0

3 years ago

Read 2 more answers

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.