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3 years ago

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:

Given information:

35, 40, 12, 16, 25, 10

Given formula:

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

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

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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 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:

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 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago

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

I need helo how to solve it. Thanks ^^