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Alenkinab [10]
3 years ago
7

A ball rolls 5.2 feet each second. How far will it roll in 1.8 seconds?

Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer:

0.3461538461538462

Step-by-step explanation:

Just divide 1.8 by 5.2. I didn't simplify it, sorry.

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x>3

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Find the area A of triangle JKL with the vertices J(5,8), K(0,7), and L(5,4)
saul85 [17]
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svp [43]

3.78 x 10^9 = 3780000000

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6 0
3 years ago
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Jessica's Furniture Store buys a couch at a wholesale price of \$113.00$113.00dollar sign, 113, point, 00.
andrew-mc [135]

Answer:

$50.85

Step by step explanation:

We have been given that Jessica's Furniture Store buys a couch at a wholesale price of $113.00 and the markup rate at Jessica's Furniture Store is 45%.

We can find mark-up for the couch by finding 45% of 113.  

\text{Mark-up}=113\times \frac{45}{100}

\text{Mark-up}=113\times 0.45

\text{Mark-up}=50.85

Therefore, mark-up for the couch will be $50.85.


4 0
3 years ago
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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
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