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Nataly [62]
3 years ago
5

Solve for this problem for N

Mathematics
2 answers:
ladessa [460]3 years ago
5 0
What in the world ? Why in the world above solve for n is two congruent triangles that appear to say a statement about them. For example, Triangle ABC is congruent to Triangle DEF. Ok idk why in the world the question has a symmetry and congruence question related to cross multiplication lol. N does equal 3 though
mezya [45]3 years ago
4 0

Answer:

N = 3

Step-by-step explanation:

I don't know what the whole thing above is, but just disregard that.

All it is here is cross multiplication.

So 28N = 21 * 4

Multiply...

28N = 84

And divide each side by 28

N = 3

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bija089 [108]

Answer:

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

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Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(score of the 36 golfers was between 70 and 71)

\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}

P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%

P(70 \leq x \leq 71) = 47.7\%

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

8 0
3 years ago
When six is subtracted from five times a number, the result is 9
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\hat{p}=0.65

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Confidence level = 0.99

We need to find the sample size to estimate the proportion of people who carry the TAS2R388 gene.

The formula for finding the sample size is:

n=\hat{p} (1-\hat{p})\left (\frac{z_{\frac{0.01}{2}}}{E} \right )^{2}

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Therefore, the required sample size is:

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Therefore, the required sample size is 3774


3 0
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