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Christopher bought new drawing pencils. He paid $8.28 for 18 pencils. At this rate, how much would it cost for 12 pencils?

Dahasolnce [82]

Answer:

$5.52

Step-by-step explanation:

The first step is finding the rate, which would be the cost per pencil. Per means for 1, so what you would do is divide $8.28 by 18.

You'll result in $0.46, or 46 cents, per pencil.

Now that you got the price per pencil, you want to multiply that by 12. This will give you the price for 12 pencils.

$0.46 x 12 = $5.52

Therefore, at this rate, it will cost $5.52 for 12 pencils.

5 0

3 years ago

Data on oxide thickness of semiconductors are as follows: 426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434,

Luda [366]

Answer:

a) 423.458

b) 9.53

c) 1.94

d) 424.5

e) 0.2916

Step-by-step explanation:

We are given the following in the question:

426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434, 435, 432, 409, 426, 409, 436, 422, 430, 411, 415

a) point estimate of the mean oxide thickness

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10163}{24} = 423.458$

b) point estimate of the standard deviation of oxide thickness

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences = 2089.958

$s = \sqrt{\dfrac{2089.958}{23}} = 9.53$

c) standard error of the point estimate

Standard error =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{9.53}{\sqrt{24}} = 1.94$

d) point estimate of the median oxide thickness

Sorted data: 409, 409, 410, 411, 412, 415, 415, 417, 420, 420, 422, 423, 426, 426, 426, 430, 430, 432, 433, 434, 434, 435, 436, 438

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

Median =

$\dfrac{12^{th}+13^{th}}{2} = \dfrac{423+426}{2} = 424.5$

e) proportion of wafers in the population that have oxide thickness greater than 430 angstrom

Sample size, n = 24

Number of wafers with oxides greater than 430 angstrom, x = 7

$p = \dfrac{x}{n} = \dfrac{7}{24} = 0.2916$

7 0

4 years ago

What is the standard deviation of the data set given below?

mixas84 [53]

Answer:

The standard deviation of the set of values (2, 3, 6, 9, 10) is square root of 12.5 ( $\sqrt{12.5}$ ) therefore OPTION C

7 0

3 years ago

Read 2 more answers

I need this answered as soon as possible

n200080 [17]

Answer:

3b to the power of 3, c to the power of 4 + 2b to the power of 4, c to the power of 3b to the power of 3, c to the power of 4, + 2b tot he power of 4, c to the power of 3

Step-by-step explanation:

sorry. hope this helps I don't know how to do exponents on my computer haha

4 0

3 years ago

What is the answer to x+2 3x-14

Mila [183]

Answer:

If your asking for x, the answer would be 8

Step-by-step explanation:

3 times 8 would be 24

24 minus 14 would be 10

and 8 plus 2 =10

10 = 10 there for x=8

4 0

3 years ago