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sergiy2304 [10]
4 years ago
11

The steps to convert 37 over 4 to a decimal are shown below

Mathematics
1 answer:
LiRa [457]4 years ago
5 0
Step 4 th eanswer is step four

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Artyom0805 [142]
no, because i have no clue the answer is false
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3 years ago
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A cartographer used a scale of 1 inch:300 meters for the heights of different mountains. If the actual height of one of those mo
lions [1.4K]

Answer:

The height on the map is of 12 inches.

Step-by-step explanation:

Scale of 1 inch:300 meters

This means that each inch on the map represents 300 meters of real height.

If the actual height of one of those mountains is 3,600 meters, what is its height on the map?​

Using proportions, by a rule of three:

1 inch - 300 meters

x inches - 3600 meters

Applying cross multiplication:

300x = 3600

x = \frac{3600}{300}

x = 12

The height on the map is of 12 inches.

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3 years ago
6 - 1 * 16 + 21 ÷ 3 4 ?
AlexFokin [52]

Answer:

18 or 9 13/34

Step-by-step explanation:

6 - 1 * 16 + 21 ÷ 3 4 ?

Remember the order of operations

6 - 1 * 16 + 21 ÷ 3/4

6 - 16 + 21  * 4/3

6 - 16 + 28

-10 + 28

18

or

6 - 1 * 16 + 21 ÷ 3 4 ?

remember the order of operations

6 - 16 + 21 ÷ 34

6 - 16 + 21/34

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9 13/34

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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
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
What does the graph sayyy
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Answer:

it says x-intercepts, multiplicity, and factors. or ar you talking about the graph on the right? well be more specific next time

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