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Romashka [77]
3 years ago
11

104 students responses 21 liked sledding as their favorite winter activity if 500 people had. Responded, how many would have bee

n expected to list sledding as their favorite winter activity? Round to the nearest whole person
Mathematics
2 answers:
lukranit [14]3 years ago
7 0
500*12/104= approximately 58 person
ankoles [38]3 years ago
4 0
The writing is confusing, but as I understand it 21 is about 20% of 104 (21x100/104). The same prosentage of 500 equals 100,96 and I would round it to 101 (the answer of the first equasion x 500/100).
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There are a total of 91 students in a drama club and a yearbook club. The drama club has 19 more students than the yearbook club
abruzzese [7]

Step-by-step explanation:

91-19=72

72/2=36

36+19=55=Drama Club students

36 =Yearbook Club Students

8 0
4 years ago
Refer to the figure to determine which statement is TRUE.
posledela

Answer:

YWZ is a right angle and XW = WZ is correct

8 0
3 years ago
A Pew Research Center report on gamers and gaming estimated that 49% of U.S. adults play video games on a computer, TV, game con
Veseljchak [2.6K]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

<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 that:

  • 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 z = 1.96.
  • 49% out of 2001 U.S. adults play video games, hence \pi = 0.49, n = 2001.

The lower bound of the interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.4681

The upper bound of the interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.5119

The 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

To learn more about the z-distribution, you can take a look at brainly.com/question/25730047

5 0
3 years ago
Sarah has 23 apples she divides them into 15 places what is the fraction of the apples she has left?
KatRina [158]
8/23 is the answer ... well that is all .. brainly asks me to write at least twenty letters so that is why i typed this much out of rage ... mark it the brainly that would make me happy when i check my phone before sleeping at night
4 0
3 years ago
Read 2 more answers
Which of the strategies below could help you find 72-25? Select all the correct answers.
irinina [24]

Step-by-step explanation:

25+5=30

30+40=70

70+2=72

8 0
3 years ago
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