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Nitella [24]
2 years ago
7

A gumball machine has 300 red gumballs. If the red gumballs are 75% of the total number of gumballs, how many gumballs are in th

e gumball machine?
225
275
375
400

BRAINLIST,EXTRA POINTS, 5 STARS, AND HEART
Mathematics
1 answer:
Arada [10]2 years ago
5 0

Answer:

400 gumballs are in the gumball machine

Step-by-step explanation:

75% = 300 gumballs

Find 1%:

1% = 300 ÷ 75 = 4 gumballs

Find 100%:

100% = 4 x 100 = 400 gumballs.

I hope this helped you! If it did, please consider rating, pressing thanks, and giving my answer 'Brainliest.' Have a great day! :)

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There are 9,481 eligible voters in a precinct. 500 were selected at random and asked to indicate whether they planned to vote fo
maria [59]

Answer:

The confidence limits for the proportion that plan to vote for the Democratic incumbent are 0.725 and 0.775.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Of the 500 surveyed, 350 said they were going to vote for the Democratic incumbent.

This means that n = 500, \pi = \frac{350}{500} = 0.75

80% confidence level

So \alpha = 0.2, z is the value of Z that has a pvalue of 1 - \frac{0.2}{2} = 0.9, so Z = 1.28.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.28\sqrt{\frac{0.75*0.25}{500}} = 0.725

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.28\sqrt{\frac{0.75*0.25}{500}} = 0.775

The confidence limits for the proportion that plan to vote for the Democratic incumbent are 0.725 and 0.775.

8 0
3 years ago
WILL MARK BRAINLEIST ​
yaroslaw [1]

Answer:

Which one do we have to answer?

Step-by-step explanation:

First of all, I can't even see the whole question. Second of all, you didn't tell us what to answer. Please comment on this answer on which question you need help with.

8 0
3 years ago
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