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ludmilkaskok [199]
3 years ago
8

Given that (-8,-8) is on the graph of f(x), find the corresponding for the function f(x)+5.

Mathematics
1 answer:
Flauer [41]3 years ago
3 0

ANSWER

(-8, - 3)

EXPLANATION

Given that: (-8,-8) is on f(x).

Then we can write (-8,f(-8)) is on f(x).

The corresponding point on

f(x) + 5

will be

(-8,f(-8)+5)

We substitute f(-8)=-8 to obtain;

(-8, - 8+5)

This simplifies to

(-8, - 3)

Therefore the corresponding point on f(x)+5 is (-8,-3).

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Step-by-step explanation:

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Translate the sentence into an equation. Seven times the difference of a number and 3 is 35.
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Read 2 more answers
Please help me, thank you!!
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A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a​ two-st
pishuonlain [190]

Answer:

a) The probability that a box containing 3 defectives will be​ shipped is 51.74\%

b) The probability that a box containing only 1 defective will be sent back for​ screening is 17.39\%

Step-by-step explanation:

Hi

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so \left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so \left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845

Finally we need to compute \frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S), therefore the probability that a box containing 3 defectives will be shipped is P(S)=54.71\%

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so \left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so \left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315

Then we need to compute \frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S), therefore the probability that a box containing 1 defective will be shipped is P(S)=82.60\%

Finally the probability that a box containing only 1 defective will be sent back for​ screening will be P(BS)=1-P(S)=1-0.8260=0.1739=17.39\%

4 0
3 years ago
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