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Natalija [7]
4 years ago
12

A students claims that 8 to the 3rd power and 8 to the negative 5th power is greater than 1. Explain whether the student is corr

ect or not.
Mathematics
1 answer:
Serga [27]4 years ago
8 0
<span>7s over 5t to the negative 3rd power 2 to the negative 3rd power times X to the 2nd power times Z to the negative 7th power 7s to the zero power times T to the negative 5th power over 2 to the negative 1st power times M to the 2nd power If you would please simplify these </span>
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Which of the numbers below belong to both the set of natural numbers and the set of integers?
VladimirAG [237]
15  -  natural (1,2,3...15....) and integers (...-2,-1,0, 1,2,...15,...)
8 0
3 years ago
Which methods correctly solve for the variable n in the equation 11−n=−20? Select all that apply. First, add n to both sides. Th
Zina [86]

Answer:

First, subtract 11 from both sides. Then, multiply both sides by −1.

First, add n to both sides. Then, add 20 to both sides.

Step-by-step explanation:

Given equation:

11 - n = -20

Solve for variable n

11 - n = -20

Subtract 11 from both sides

11 - n - 11 = -20 - 11

- n = - 31

Multiply both sides by -1

- n(-1) = - 31(-1)

n = 31

First, subtract 11 from both sides. Then, multiply both sides by −1.

11 - n = -20

11 -n +n = -20 + n

11 = -20 + n

11 + 20 = -20 + n +20

31 = n

First, add n to both sides. Then, add 20 to both sides.

First, subtract n from both sides. Then, multiply both sides by −20.

First, add n to both sides. Then, subtract 11 from both sides.

11 - n = -20

11 - n +n = -20 + n

11 = -20 + n

11 - 11 = -20 + n -11

Not possible

First, subtract 11 from both sides. Then, multiply both sides by −1.

First, add n to both sides. Then, subtract 20 from both sides.

11 - n = -20

11 - n + n = -20 + n

11 = -20 + n

11 - 20 = -20 + n - 20

9 = n - 40

Not possible

5 0
3 years ago
Given the function F(x), you can get a picture of the graph of its inverse F -1(x) by flipping the original graph of F(x) over t
Lana71 [14]
C=1 because you are flipping it over the line so it cant be negative
8 0
3 years ago
Read 2 more answers
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
3 years ago
Can yall pls help with this
Blizzard [7]

Answer:

x=7

Step-by-step explanation:

3 0
3 years ago
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