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Gelneren [198K]
2 years ago
9

HELP ASAP!!!!!!!!!!!!!!!

Mathematics
2 answers:
Dafna11 [192]2 years ago
8 0

Answer:

38 degrees

Step-by-step explanation:

Since angles KLO and OLM are supplementary, they add up to 180 degrees

So,  angle OLM is 180-134=46 degrees

Similarly, angles OMN and OML are supplementary, so they also add up to 180 degrees

So, angle OML is 180-84=96 degrees

You also know that the angles of a triangle sum up to 180 degrees

So, 46+96+x=180

180-142=38

x=38 degrees

charle [14.2K]2 years ago
7 0
Angles on a straight line sum up to 180degrees
So 180-134= 46 and 180-84=96
So interior angles of a triangle sum up to 180
Hence x + 46 + 96 =180
X+142=180
X= 180-142
X=38
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3 years ago
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)

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           \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 .

                                                                             

8 0
3 years ago
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CaHeK987 [17]

Answer:

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5 0
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