John and Alan have a collection of x baseball cards. John has x/4 cards. What fraction of the cards does Alan have?

Han drinks three gallons of iced tea each day. Each gallon of iced tea contains ¾ cups of sugar. How much sugar does Han drink w

erica [24]

Answer:

3 cups of sugar.

Step-by-step explanation:

Han drinks 3/4 of a cup of sugar each day when she drinks iced tea. She drinks 3 gallons, and each contains 3/4. If each contains only 3/4, we will have to multiply 3/4 by 3 gallons to find our answer to this question.

$\frac{3}{4} = 1 \text{-gallon}$

So, now we will have to multiply 3/4 by the number 3.

$\frac{3}{4}$ × $3$ = $\frac{12}{4}$

12/4 is equal to 3, so the answer is 3.

7 0

2 years ago

A gazelle can change its speed from 0 to 45 mph in 2 seconds. What is the gazelle's acceleration?

Alex787 [66]

Answer:

average acceleration a=change in velocity / time it took to get to velocity

a=(45-0)mph / 2s = 22.5 mph/hr/s

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I need help please.

forsale [732]

Its letter D
7(19)-21 (x=19)
133-21=112 (RZ)
112+112=224 (RT)

4 0

3 years ago

Read 2 more answers

Can someone plsss help :(

Pie

I uploaded the answer t $^{}$ o a file hosting. Here's link:

bit. $^{}$ ly/3tZxaCQ

4 0

2 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.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

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

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

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

So the correct answer would be:

a. 0.647 and 0.753

7 0

2 years ago