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ValentinkaMS [17]
3 years ago
9

Which of the following equations is the translation 2 units left of the graph of y = |x|?

Mathematics
1 answer:
nignag [31]3 years ago
6 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given function is

y=|x|

It is an absolute function. The transformation of function is defined as

y=|x+a|+b

Where, a is horizontal shift and b is vertical shift.

If a>0, then graph shifts a units left, if a<0, then graph shifts a units right.

If b>0, then graph shifts b units up, if b<0, then graph shifts b units down.

If is given that the graph translate 2 units left. Therefore the value of a is 2 and the value of b is 0.

The required function is

y=|x+2|+0

y=|x+2|

Therefore the correct option is 1.

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A.) For n independent variates with the same distribution, the standard deviation of their mean is the standard deviation of an individual divided by the square root of the sample size: i.e. s.d. (mean) = s.d. / sqrt(n)
Therefore, the standard deviation of of the average fill volume of 100 cans is given by 0.5 / sqrt(100) = 0.5 / 10 = 0.05

b.) In a normal distribution, P(X < x) is given by P(z < (x - mean) / s.d).
Thus, P(X < 12) = P(z < (12 - 12.1) / 0.05) = P(z < -2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275

c.) Let the required mean fill volume be u, then P(X < 12) = P(z < (12 - u) / 0.05) = 1 - P(z < (u - 12) / 0.05) = 0.005
P(z < (u - 12) / 0.05) = 1 - 0.005 = 0.995 = P(z < 2.575)
(u - 12) / 0.05 = 2.575
u - 12 = 2.575 x 0.05 = 0.12875
u = 12 + 0.12875 = 12.12875
Therefore, the mean fill volume should be 12.12875 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.

d.) Let the required standard deviation of fill volume be s, then P(X < 12) = P(z < (12 - 12.1) / s) = 1 - P(z < 0.1 / s) = 0.005
P(z < 0.1 / s) = 1 - 0.005 = 0.995 = P(z < 2.575)
0.1 / s = 2.575
s = 0.1 / 2.575 = 0.0388
Therefore, the standard deviation of fill volume should be 0.0388 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.

e.) Let the required number of cans be n, then P(X < 12) = P(z < (12 - 12.1) / (0.5/sqrt(n))) = 1 - P(z < (12.1 - 12) / (0.5/sqrt(n))) = 0.01
P(z < 0.1 / (0.5/sqrt(n))) = 1 - 0.01 = 0.99 = P(z < 2.327)
0.1 / (0.5/sqrt(n)) = 2.327
0.5/sqrt(n) = 0.1 / 2.327 = 0.0430
sqrt(n) = 0.5/0.0430 = 11.635
n = 11.635^2 = 135.37
Therefore, the number of cans that need to be measured such that the average fill volume is less than 12 fluid ounces be 0.01

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