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3 years ago

All of the following expressions are equal except ____. 1/4^3 4^2/4^5 4^4/4^2 4^-3

Mathematics

2 answers:

Bas_tet [7]3 years ago

5 0

1/4^3= 0.015625

4^2/4^5= 0.015625

4^4/4^2= 16

4^-3= 0.015625

therefore the third one is the one that u are looking for

4^4/4^2= 16

hram777 [196]3 years ago

3 0

Answer:

the answer is 4^4/4^2

Step-by-step explanation:

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3 years ago

Anumeha is mowing lawns for a summer job. For every mowing job, she charges an initial fee of $ 10 $10dollar sign, 10 plus a con

slavikrds [6]

Answer:

F(t) = 10 + 5(t)

Step-by-step explanation:

The complete question is as follows;

Anumeha is mowing lawns for a summer job. for every mowing job, she charges an initial fee of \$10$10dollar sign, 10 plus a constant fee for each hour of work. her fee for a 555-hour job, for instance, is \$35$35dollar sign, 35. let f(t)f(t)f, left parenthesis, t, right parenthesis denote anumeha's fee for a single job fff (measured in dollars) as a function of the number of hours ttt it took her to complete it. write the function's formula.

Solution

We are interested in writing the function F(t) formula for the fee charged by Anumeha per job.

Now, the key to writing this function is knowing exactly the constant fee she charges on the job.

We were told that she got $35 for a 5 hour job.

Thus, the constant amount charged is as follows;

Since it’s $10 as initial fee and the constant fee is per hour;

35 = 10 + 5(x)

where x is the constant fee per hour

35 = 10 + 5x

5x = 35-10

5x = 25

x = 25/5

x = $5

This mess that she charges a constant fee of $5 per hour

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F(t) = 10 + 5(t)

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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.

3 0

3 years ago