Write a word problem that could be solved with the equation<br> 0.15x + 200 650

Soft-drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounces assume that the fill v

evablogger [386]

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

8 0

3 years ago

Katie has 1/6 of watermelon left over from her picnic that she wants to split between her mom and dad. How much watermelon will

frozen [14]

Answer: 1/12 of a watermelon

Each parent will get one half of 1/6 each.
1/2 x 1/6 = 1/12

3 0

4 years ago

A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

$g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)$

Then, first and second derivatives of the function are, respectively:

First derivative

$g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)$

$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

$g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x$

$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

$g''(x) = 12\cdot x^{2}-30\cdot x -28$

Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

$g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28$

$g''(0) = -28$

This value leads to a local maximum.

x = 5.118

$g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28$

$g''(5.118) = 132.787$

This value leads to a local minimum.

x = -1.368

$g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28$

$g''(-1.368) = 35.497$

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

$g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}$

$g(0) = 0$

Local minima

$g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}$

$g(5.118) = -350.90$

$g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}$

$g(-1.368) = -9.90$

7 0

4 years ago

What is the area of 2 2/5 and 4 1/8

KengaRu [80]

The area is 9.9 or 9 9/10

4 0

3 years ago

Explain how you know that 7/12 is greater than 1/3 but less than 2/3

PtichkaEL [24]

Or this question you should know that:
1/3 = (1*4)/(3*4) = 4/12
and
2/3 = (2*4)/(3*4) = 8/12
so you can see that 7/12 is greater than 4/12 which is 1/3 and less than 8/12 which is 2/3 :))))
i hope this is helpful
have a nice day

7 0

3 years ago

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Write a word problem that could be solved with the equation 0.15x + 200 650