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3 years ago

14

A shoebox is covered with packing paper before being shipped. The paper cost $0.02 per square inch. How much will it cost to cov

er three pairs of shoes?

1 answer:

Yuliya22 [10]3 years ago

3 0

It will cost about $0.72 for three pair of shoes assuming shoe box is rectangular shape having 12 square inches of packing paper.

Step-by-step explanation:

Assuming a shoe box rectangle in shape and the same size is required.
Length of rectangle box assuming to be 4 inches.
Width of rectangle box assuming to be 3 inches.
as we know area of rectangle is Length * Breath(Width) .
12 square inches is the total packing paper required for each box.
To cover cost of three pairs of shoes $0.02 per square inch.
Each box would cost about 12 square inches * 0.02 pr square inches.
0.24 square inches of packing paper will be required.
let three pairs of shoes be x
Total pairs of shoes would be= (12 * 0.02)*x
Answer would be for 3 pairs of shoes = 0.24*3 = 0.72 square inches.

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In a parallel circuit et= 240 v ,r = 30 ohms and xl = 40 ohms what is the power factor

SVEN [57.7K]

Answer:

$PF=2400$ kVA units.

Step-by-step explanation:

Given : In a parallel circuit et= 240 v ,r = 30 ohms and xl = 40 ohms.

To find : What is the power factor?

Solution :

In a parallel rl circuit we have,

Total voltage, et= 240 Volts,

Resistance, r= 30 ohm,

Inductive Reactance, xl= 40 ohm.

In parallel circuit, the voltage across resistor and inductor remains same.

i.e. vr = vl= v= 240 Volts

Using Ohm's Law,

The current across resistor would be $ir=\frac{v}{r}$

$ir=\frac{240}{30}$

$ir=8\ amp$

Using Ohm's Law,

The current across inductor would be $il=\frac{v}{xl}$

$il=\frac{240}{40}$

$il=6\ amp$

Now, Total current would be $(it)^2= (ir)^2+ (il)^2$

$(it)^2= (8)^2+ (6)^2$

$(it)^2= 64+36$

$(it)^2= 100$

$it= 10$

The voltage is 240 V and current is A= 10 amp

The power factor is $PF=V\times A$

$PF=240\times 10$

$PF=2400$ kVA units.

4 0

4 years ago

Juanita is saving money for a car. The vehicle she wants costs 7,500 dollars. A relative gave her part of the money she needs be

grin007 [14]

80 * 42 gives you how much she needs to earn = 3360

7,500 - 3360 = how much her relative gave her = 4140

3 0

3 years ago

Read 2 more answers

What is 4pie=w-6 pie

Marina CMI [18]

W= 31.4
Step 1: 4*Pi = 12.566370614359
Step 2: -6*Pi = -18.849555921539

Step 3: Rewrite equation (12.566370614359=w-18.849555921539)
Step 4: Move the number that is equivalent to 6Pi over to the other side of the equal sign to get the variable alone. (12.566370614359+18.849555921539 = 31.415926535898 ) The problem should now look like this, <span>31.415926535898=w

Step 5: Divide the numbers (which you don't really need to do here since the "number" in front of w is 1. This is helpful to know for future problems though.)

Step 6: Round to the nearest tenths place (or however you need to) giving you the final answer of 31.4 </span>

4 0

4 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

Use the Distributive Property to simplify 5/6(x + 1/2 + 4)

zaharov [31]

5/6(x+1/2+4)

~Multiply

5/6 * x/6 + 5/6 * 1/6

~Simplify

5/36x + 5/36

Best of Luck!

8 0

3 years ago