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Ivenika [448]
3 years ago
6

Please help me, I need answers in 2 days. Please give answers in one picture otherwise i cannot view it without paying. Would be

very appreciated. Will give branliest

Mathematics
1 answer:
solniwko [45]3 years ago
6 0

Answer:

Step-by-step explanation:

i hope this is good enough

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Please help!!!!this is my last part of my discussion and I am stuck on it I would really appreciate the help!!
frosja888 [35]

x + 5 = 0 - > x = -5

5 0
3 years ago
There is a decrease in power. It can be concluded that there has also been a change in: A)current or voltage.
Serhud [2]
There is a decrease in power. It can be concluded that there has also been a change in:

A) current or voltage.
5 0
3 years ago
The volume of the cone is 130 units^3. Solve for x.<br> Height: 13<br> Radius: x
Gwar [14]
V=(1/3)hpir^2
v=130
h=13
r=x

130=(1/3)(13)pix^2
times both sides by 3
390=13pix^2
divide both sides by 13
30=pix^2
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3 0
3 years ago
The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab
Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

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}.

The margin of error is:

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

For this problem, we have that:

\pi = 0.25

99% confidence level

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.

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when M = 0.1

So

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

0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}

0.1\sqrt{n} = 2.575{0.25*0.75}

\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}

(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}

n = 124.32

Rounding up

The minimum sample size needed is 125.

5 0
4 years ago
Pls help me so I cant be done with school'
Alenkinab [10]

Answer:

2 2/5

Step-by-step explanation:

8 0
3 years ago
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