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Westkost [7]
3 years ago
15

PLEASE HELP ME!!!! thank u

Mathematics
2 answers:
Romashka [77]3 years ago
8 0
<h3>Answers:</h3>
  • The equation is C = 1500S + 4500
  • The graph is a straight line that goes through (0,4500) and (1,6000)

================================================================

Explanation:

S = number of tons of sugar

150S = cost of transporting all S tons of sugar

150S+4500 = total cost since we add on the $4500 fixed cost

C = 150S+4500 is the cost equation

--------------------------

If we replace C and S with y and x (in that order), then the equation turns into y = 1500x+4500

To graph this, you only need two points since this is a linear equation in the form y = mx+b. We have m = 1500 as the slope and b = 4500 as the y intercept.

To find a point, you plug in some number for x to get a corresponding y value. For instance, if x = 0

y = 1500x+4500

y = 1500(0)+4500

y = 4500

Showing (0,4500) is the y intercept and one point on the line

Now plug in x = 1

y = 1500x+4500

y = 1500(1)+4500

y = 6000

Showing (1,6000) is another point on the line

You can keep going to generate more points, but two is enough.

The graph will be a straight line through (0,4500) and (1,6000)

tester [92]3 years ago
5 0

Answer:

Step-by-step explanation:

The answer is (-26,3)

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8 0
4 years ago
Twenty-five samples of 100 items each were inspected when a process was considered to be operating satisfactorily. In the 25 sam
notsponge [240]

a) The estimate of the proportion of defectives when the process is in control is 0.054

b) The standard error of the proportion if the sample size is 100 is 0.0226.

c) The upper control limit is 0.1218 and the lower control limit is 0 (since LCL < 0 and p > 0, we can write LCL = 0).

<h3>What are the formulas for finding the estimate of the proportion, standard variation, and control limits?</h3>

1) The estimate of the proportion of success is

p = (number of success)/(total number of samples)

I.e., p = x/N

2) The standard deviation of the proportion of success is

\sigma_p = \sqrt{\frac{p(1-p)}{n} }

3) The upper and lower control limits for a control chart are:

L.C.L = p - 3\sigma_p

and U.C.L = p + 3\sigma_p

<h3>Calculation:</h3>

It is given that, there are 25 samples of 100 items each.

So, the total number of items i.e., the total sample size,

N = 25 × 100 = 2500

In 25 samples, a total of 135 items were found to be defective.

So, the number of defectives x = 135

a) The estimate of the proportion of defectives is p = x/N

On substituting, we get

p = 135/2500 = 0.054

b) The standard error of the proportion if the sample of size 100 is calculated by

\sigma_p = \sqrt{\frac{p(1-p)}{n} }

On substituting p = 0.054 and  n = 100, we get

\sigma_p = \sqrt{\frac{0.054(1-0.54)}{100} }

    = 0.0226

c) The control limits for the control chart are:

Upper control limit =  p + 3\sigma_p

⇒ U.C.L = 0.054 + 3(0.0226) = 0.054 + 0.0678 = 0.1218

Lower control limit = p - 3\sigma_p

⇒ L.C.L = 0.054 - 3(0.0226) = 0.054 - 0.0678 = - 0.0138 ≈ 0

(Since we know that the lower control limit should not be a negative value, it is made equal to 0).

Learn more about an estimate of the proportion here:

brainly.com/question/23986522

#SPJ4

7 0
1 year ago
Please help! will give brainliest. click for the picture
vesna_86 [32]

Answer:

Gimme some time

Step-by-step explanation:

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3 years ago
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