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grigory [225]
2 years ago
11

HELP NOWW PLEASE !!!!!

Mathematics
2 answers:
Sergeu [11.5K]2 years ago
4 0

Answer:

c

Step-by-step explanation:

nirvana33 [79]2 years ago
4 0

Answer:

e

Step-by-step explanation:

8x10

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The perimeter of the warehouse can be modeled by which of the following? P = 4(348 feet) P = (348 feet)2 P = 2(348 feet) + 2(348
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P= 4(348) because assuming the wear house has four sides that is the only logical answer
3 0
3 years ago
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Prove this trigonometric equation;<br><br> - tan^2x + sec^2x = 1,
alekssr [168]
Hey there :)

- tan²x + sec²x = 1    or    1 + tan²x = sec²x

sin²x + cos²x = 1 
Divide the whole by cos²x

\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

\frac{sinx}{cosx} = tanx so \frac{sin^2x}{cos^2x} = tan^2x
and
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Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}

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3 0
2 years ago
You have learned that different types of rocks are found in Earth’s layers. Their composition affects their densities.
gtnhenbr [62]

Answer:

c the number of the person who is this and what is the address of the house and the

Step-by-step explanation:

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6 0
3 years ago
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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

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