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3 years ago

Solve for t.

Mathematics

2 answers:

finlep [7]3 years ago

7 0

Your answer would be A)

Hope this helps (:

Tpy6a [65]3 years ago

4 0

Answer:

A( -8

Step-by-step explanation:

i took the test and got 100% ;)

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I will gave brainestly

hammer [34]

Answer:

Hello !

The solutions are: -8 , -4 , 6

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3 0

3 years ago

Need answer Like ASAP

kogti [31]

Answer:

I believe its C

Step-by-step explanation:

My bad if im wrong tho

8 0

3 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!!! CAN you help there is a website it has the answers appealy but I cannot seem to find them

miv72 [106K]

Answer:

The difference is While pay grades are administrative classifications used primarily to standardize compensation across the military services, ranks indicate a level of responsibility (for personnel, equipment, and mission) which grows with each increase in rank.

Step-by-step explanation:

There similar because Pay grades E-1 through E-3 generally indicate basic training, advanced training or entry-level work. For the Army and Marines, E-1 corresponds to the rank of private

I hoped this helped!

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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

Find the length of the midsegment.

STALIN [3.7K]

Answer:

Step-by-step explanation:

The length of the midsegment is half the length of the base segment, so you have ...

... 2(6x +2) = 2x +84

... 10x = 80 . . . . . . . . . simplify, subtract 2x+4

... x = 8

The length of the midsegment is 6·8+2 = 50.

7 0

3 years ago