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mote1985 [20]
3 years ago
14

What are the first two answers? #16 and #17

Mathematics
1 answer:
Luden [163]3 years ago
5 0
Answer to number 16 is 27 and 17 is 2
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What is the postulate?
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Suggest or assume the existence, fact, or truth of something as a basis for reasoning, discussion, or belief.
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Linda and Juan went shopping. Linda spent $14 less than Juan.
Rudiy27

Answer: Let x = amount Linda spent.

x=y-$18

Step-by-step explanation:

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3 years ago
The area of a square is 50 meters. how long is each side? round your answer to the nearest meter.
DanielleElmas [232]
The closest would be 7
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3 years ago
Read 2 more answers
Plz help me thanks very nuch whoever helped me​
photoshop1234 [79]

Answer:

(1) 56 miles/hour

Step-by-step explanation:

We need to find the average rate of change from t = 2 to t = 9.

At t = 2 hours, d = 106 miles.

At t = 9 hours, d = 498 miles.

The average rate of change in function f(x) from x = a to x = b is

[f(b) - f(a)]/(b - a)

average rate of change from t = 2 to t = 9 =

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Answer: (1) 56 miles/hour

7 0
2 years ago
It has a time to failure distribution which is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehi
Triss [41]

Answer:

The designed life should be of 21,840 vehicle miles.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.

This means that \mu = 35000, \sigma = 7000

Find its designed life if a .97 reliability is desired.

The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.

Z = \frac{X - \mu}{\sigma}

-1.88 = \frac{X - 35000}{7000}

X - 35000 = -1.88*7000

X = 21840

The designed life should be of 21,840 vehicle miles.

3 0
2 years ago
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