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3 years ago

Which angles are acute? check all that apply.

Mathematics

2 answers:

Montano1993 [528]3 years ago

5 0

Right is acute like the answer above this one

Alex17521 [72]3 years ago

3 0

Right us acute like leebron jameson

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WILL MARK BRAINLIEST solve for x: m(2m-x)=3(x+6) if m≠3

Lera25 [3.4K]

Answer:

x=2m−6

Step-by-step explanation:

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3 years ago

Which expression is equivalent?

Irina18 [472]

Answer:

Again! Option D is the answer!

Please mark my answer as the brainliest since I have you the answer so quickly!

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3 years ago

The Candy Shack sells 138 pounds of candy on Tuesday. They sold 71.76 pounds of jelly beans. What percent of the candy sold was

Pepsi [2]

Answer:

<h2>52% were jelly beans!</h2>

Explanation:

Cross multiply the following:

$\frac{x}{100} = \frac{71.76}{138}$

x times 138 = 138x

100 times 71.76 = 7,176

Divide both sides by 138:

$\frac{138x}{138} = \frac{7176}{138}$

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3 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 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:

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

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3 years ago

What is .25 when written in simplest form

seropon [69]

So 0.25 is essentially 25/100. Divide 25 from the top and bottom to get 1/4

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3 years ago

Read 2 more answers