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3 years ago

PLEASE HELP WITH #4

Mathematics

1 answer:

Phantasy [73]3 years ago

7 0

A) x=1,-13
B) x= -1 +- square root of 21 dvided by 10
C) x= 3 +- i square root of 7 divided by 2
D) x= 1 +- i square root of 11 divided by 4

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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, a

Bas_tet [7]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

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 interval $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.

4 0

3 years ago

Direction: Arrange the following numbers in decreasing order. Write your answer on the blank provided below each number.

rjkz [21]

Answer:

1. 100 000, 89 000, 67 000, 55 000

2. 45 678, 34 567, 23 456, 12 345

3. 98 765, 87 654, 76 543, 65 432-

7 0

3 years ago

Multiply 3xyz ^ 2 / 6y ^ 4 by 2y / x z ^ 4

Marat540 [252]

$\dfrac{3xyz^2}{6y^4}=\dfrac{3}{6}\cdot\dfrac{xyz^2}{y^4}=\dfrac{1}{2}\cdot\dfrac{xz}{y^3}=\dfrac{xz}{2y^3}\\\\\\\dfrac{3xyz^2}{6y^4}\cdot\dfrac{2y}{xz^4}=\dfrac{xz^2}{2y^3}\cdot\dfrac{2y}{xz^4}=\dfrac{2}{2}\cdot\dfrac{xyz^2}{xy^3z^4}=1\cdot\dfrac{1}{y^2z^2}=\dfrac{1}{y^2z^2}$

8 0

3 years ago

Read 2 more answers

30km=?miles 44,000in=?cm 21ft=?m 0.75miles=?dam 133yards=mm Please show your work

kherson [118]

So there is this chart my teacher showed me to help find the answers to your questions I will send a picture but here are the answers

30km is 18.642m
44,000in is 111760cm
21ft is 6.401m
0.75m I’m not for sure sorry but
133yards is 121615.2mm

I hope that helps sorry I did not understand the one question 0.75m

5 0

3 years ago

Square root of 214 show work thanks

Nataliya [291]

√214= 14.6287388383278

Calculator maybe??

6 0

3 years ago