HELP ASAPPPPPPP!!!!!!!!!!!!!!!50 POINTS!!!!!!!!

Mathematics

1 answer:

Ratling [72]2 years ago

3 0

I have no internet for the past week and it won’t work on me lol I don’t have internet to say I

You might be interested in

A pair of parallel lines is cut by a transversal as shown in the figure. Match the angles with their corresponding angles.

garri49 [273]

D and H are corresponding
G and C are corresponding
E and A are corresponding
B and F are corresponding

4 0

2 years ago

Read 2 more answers

Jan takes an IQ test at ages 7 and 9. Mike takes an IQ test at ages 5 and 9. Mark takes an IQ test at ages 7 and 10. Jane takes

sasho [114]

The IQ scores are most likely to be more positively correlated is Jan

7 0

1 year ago

URGENT

nadya68 [22]

Answer:

Scalene, Acute

Step-by-step explanation:

Scalene because each side has a different length

obtuse because one angle is more than 90°

6 0

2 years ago

I don’t really understand this

mars1129 [50]

Uhh me neither thats very complicated

7 0

2 years ago

Read 2 more answers

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

2 years ago