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3 years ago

15

Please explain this subject to me with work shown use 3.14 for pie please explain. ASAP If you do answer a question please say w

hich number 22 points and Brainliest

1 answer:

podryga [215]3 years ago

7 0

Answer: V=718.378

Step-by-step explanation:

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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were born, a

Alex Ar [27]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

In the study 380 babies were born, and 342 of them were girls.

This means that $n = 380, \pi = \frac{342}{380} = 0.9$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.8604$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.9396$

As percentages:

0.8604*100% = 86.04%.

0.9396*100% = 93.96%.

The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.

4 0

3 years ago

(9.5•10/11)+(6.3•10/9

maria [59]

15 7/11 I think im not sure

5 0

3 years ago

Read 2 more answers

Factor completely. y^6 - 49x^6

HACTEHA [7]

Answer:

(y+7x^3) (y-7x^3)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

15 points! The data set shows the weights of pumpkins, in pounds, that are chosen for a photograph in a farming magazine. The re

vladimir2022 [97]

Answer:

Step-by-step explanation:

a). Mean of the given data set = $\frac{32+28+18+40+22}{5}$

= 28

b). x $x-\overline{x}$ $(x-\overline{x})^2$

32 32-28 = 4 16

28 28 - 28 = 0 0

18 18 - 28 = -10 100

40 40 - 28 = 12 144

22 22 - 28 = -6 36

$\sum({x-\overline x})^2$ = 296

c). Standard deviation = $\sqrt{\frac{\sum (x-\overline{x})^2}{n}}$

= $\sqrt{\frac{296}{5} }$

= 7.69

d). Mean of the new set of weights = $\frac{32+33+34+35+36}{5}$

= 34

Since mean of the data set is more close to each data, standard deviation will be less than the deviation given in part (c).

7 0

3 years ago

Bas_tet [7]

Huh english is this spanish !

7 0

3 years ago