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Vladimir79 [104]
3 years ago
10

Hey! There is a link and I need help with these 3 problems for math. Pls give me awnsers and I will mark brainlist, have a good

day!!

Mathematics
2 answers:
Soloha48 [4]3 years ago
8 0
1: 9x+3
2: 3m^2+9m-7
3: 3x^2+3y^2+y+4

Those are all of the answers in order! I hope this helps.
marissa [1.9K]3 years ago
4 0

okay so basically heres what i got

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The data given to the right includes data from ​candies, and of them are red. The company that makes the candy claims that ​% of
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Using the z-distribution, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

Researching this problem on the internet, 8 out of 39 candies are red, hence the sample size and the estimate are given by:

n = 39, \pi = \frac{8}{39} = 0.2051

Hence the bounds of the interval are:

  • \pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 - 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.0784
  • \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 + 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.3318

As a percentage, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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