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wariber [46]
3 years ago
6

Find the inverse of the function. f(x) = 7x + 2

Mathematics
1 answer:
solmaris [256]3 years ago
3 0
F ( x ) = 7 x + 2
y = 7 x + 2
- 7 x = - y + 2  /*(-1)
7 x = y - 2   / : 7
x = y/7 - 2/7 = ( y - 2 ) / 7
The inverse of the function is:
f^(-1) ( x ) = ( x - 2 ) /7 

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Combine any like terms in the expression. If there are no like terms rewrite the expression.
swat32

Answer:

0

Step-by-step explanation:

7jk + 3jk = 10jk

10jk - 4jk = 6jk

6jk - 6jk = 0jk

0jk = 0

4 0
3 years ago
9 is .03% of what number?
iren2701 [21]

Answer:

30,000

Step-by-step explanation:

We know that divide the percentage by 100. After that you get the decimal thing then all you do is multiply the number and you get your answer that is  30,000.

Answer: 30,000

Please mark brainliest

<em><u>Hope this helps.</u></em>

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3 years ago
1.find the length of the side indicated. And round to 1 decimal place
vesna_86 [32]

Answer:

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3 years ago
A professor gives his students 9 essay questions to prepare for an exam. Only 3 of the questions will actually appear on the exa
Talja [164]

Answer: 84

Step-by-step explanation:

The total number of the questions is 9

Possible question is 3

This is combination

Therefore n=9 and r=3

Combination formula=n! / ((n-r)! r!)

9C3=9! /((9-3)! 3!)

=362880 /(6! 3!)

=362880/4320

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4 0
3 years ago
In recent years, only 79% of the fruit sold at Fred's Fruit stand has been any good. Fred recently started buying fruit from new
Juli2301 [7.4K]

Answer:

The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

Step-by-step explanation:

79% of the fruit sold at Fred's Fruit stand has been any good. Test if there has been an increase.

At the null hypothesis, we test if there has been no increase, that is, the proportion is still of 79%, so:

H_0: p = 0.79

At the alternative hypothesis, we test if there has been an increase, that is, more than 79% being good, so:

H_1: p > 0.79

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.79 is tested at the null hypothesis:

This means that \mu = 0.79, \sigma = \sqrt{0.79*0.21}

In a random sample if 475 pieces, 85 were bad.

So 475 - 85 = 390 were good, and:

n = 475, X = \frac{390}{475} = 0.8211

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.8211 - 0.79}{\frac{\sqrt{0.79*0.21}}{\sqrt{475}}}

z = 1.66

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.8211, which is 1 subtracted by the p-value of Z = 1.66.

Looking at the z-table, z = 1.66 has a p-value of 0.9515.

1 - 0.9515 = 0.0485

The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

5 0
3 years ago
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