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Lelu [443]
3 years ago
6

john is now three times as old as his brother Sam. In five years, john will be twice as old as Sam will be then. Find their pres

ent ages.
Mathematics
1 answer:
shtirl [24]3 years ago
4 0
J=john's age now
s=sam's age now

j=3 times s
j=3s
in five years (j+5,s+5)
j+5=2 times (5+s)
j+5=2(s+5)
j+5=2s+10
subtract 5 from both sides
j=2s+5
use other equation
j=3s
j=2s+5
subsitute
3s=2s+5
subtract 2s
s=5

sam=5
j=3s
j=3 times 5
j=15


john=15
sam=5
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Answer:

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Which expression is equivalent to this expression? 8(11y - 5)
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3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and
lisov135 [29]

Answer:

(a) 0.50928

(b) 0.857685.

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 155 lb and a standard deviation of 29.2 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 - 155}{29.2} ) = P(Z < 1.57) = 0.94179

P(X <= 150) = P( \frac{X - \mu}{\sigma}  < \frac{150 - 155}{29.2} ) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749

                                                                                                   = 0.43251

Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .

(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 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < 9.84) = 1 - P(Z >= 9.84)

                                                                                  = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < -1.07) = 1 - P(Z < 1.07)

                                                                                   = 1 - 0.85769 = 0.14231

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.

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.

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3 years ago
Hey.. I’m really stuck on this page of homework, can anyone help?
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Answer:

Easy!

Step-by-step explanation:

All you have to do is..       Percentage = (number you want to find the percentage for ÷ total) × 100. Move the decimal point two places to the right to convert from a decimal to a percentage, and two places to the left to convert from a percentage to a decimal.

Hope this helps :)

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