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chubhunter [2.5K]
3 years ago
6

I need helo how to solve it. Thanks ^^​

Mathematics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

42.5

Step-by-step explanation:

It equals 90 degrees because it's a right angle, so:

3 + 2 = 5

90 - 5 = 85

Since there is space at the top, you need to have the correct amount for both sides. So:

85 ÷ 2 = 42.5

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Find the mean 35 40 12 16 25 10
ruslelena [56]

Answer: Mean = 23

Step-by-step explanation:

<u>Given information:</u>

35, 40, 12, 16, 25, 10

<u>Given formula:</u>

Mean~=~\frac{Sum~of~terms}{Number~of~terms}

<u>Substitute values into the formula</u>

Mean~=~\frac{35~+~40~+~12~+~16~+~25~+~10}{6}

<u>Combine like terms</u>

Mean~=~\frac{75~+~28~+~35}{6}

Mean~=~\frac{103~+~35}{6}

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\Large\boxed{Mean=23}

Hope this helps!! :)

Please let me know if you have any questions

4 0
2 years ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
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