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2 years ago

1.4.

Mathematics

1 answer:

Ann [662]2 years ago

6 0

Answer:

need more details

Step-by-step explanation:

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Someone help me with this geometry homework please

marshall27 [118]

Look at the attached files.

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3 years ago

A softball team is ordering pizza to eat after their tournament. They plan to order cheese pizzas that cost $ 6 each and four-to

Vlada [557]

Answer:

6x + 10t

Step-by-step explanation:

We know that each cheese pizza (c) is $6, and each four topping pizza (f) is $10.

Knowing this, the total of all the pizzas is going to be 6* the amount of cheese (c) plus 10* the amount of four-toppings.

6x + 10t

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3 years ago

Sam owes the bank a little less then $40 what could be his balance

bija089 [108]

D 39.25 which isn't far from 40

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3 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.