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forsale [732]
4 years ago
9

Find the area of the shaded triangle and show each step and give your answer using the correct units

Mathematics
1 answer:
Svetllana [295]4 years ago
3 0
Area=26x17=442




The answer is 442
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Solving F=ma for m:  m=F/a.  Thus, if F=25 units and a=5 units, m=25/5 units, or m=5 units.
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Mary lost 14 pounds, which is 10% of her starting weight. What was her original weight?
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Solve for y:-<br><br> y - 18 = 10<br><br> y = ? <br><br> Thank You!
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3 years ago
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An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and
charle [14.2K]

Answer:

A) 0.5737

B) 0.9884

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane.  The new population of pilots has normally distributed weights with a mean of 160 lb and a standard deviation of 27.5 lb i.e.;                                                 \mu = 160 lb  and \sigma = 27.5 lb

(A) We know that Z = \frac{X - \mu}{\sigma} ~ N(0,1)

Let X = randomly selected pilot  

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( \frac{X - \mu}{\sigma} < \frac{201 - 160}{27.5} ) = P(Z < 1.49) = 0.9319

P(X <= 150) = P( \frac{X - \mu}{\sigma} < \frac{150 - 160}{27.5} ) = P(Z < -0.3636) = P(Z > 0.3636) = 0.3582

Therefore, P(150 < X < 201) = 0.9319 - 0.3582 = 0.5737 .

(B) We know that for sampling mean distribution;

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

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

 P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

  P(X bar < 201) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{201 - 160}{\frac{27.5}{\sqrt{39} } } ) = P(Z < 9.311) = 1 - P(Z >= 9.311)

                                                                                    = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 160}{\frac{27.5}{\sqrt{39} } } ) = P(Z < -2.2709) = P(Z > 2.2709)

                                                                                          = 0.0116

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.0116 = 0.9884

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

8 0
3 years ago
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