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3 years ago

Pls help meh pls Find the area of the figure. (Sides meet at right angles.)

Mathematics

1 answer:

Vlad [161]3 years ago

3 0

We find the area of the total square which is 8in • 8in=64in^2 . Now we find the area of the small quadrilateral which is 4in•4in=16 in^2.
Now we subtract the area of the small quadrilateral from the big quadrilateral’s area 64in^2 -16in^2 leaving us with the are of 48in^2 of the figure

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What is the circumference (in inches) of a circle with radius of 5 inches? Use 3.14 for n.

Yuliya22 [10]

Answer:

A. 31.4 in.

Step-by-step explanation:

Formula for the circumference:

C = dπ (d = 2r)

Therefore:

C = 10π

C = 10 · 3.14

C = 31.4 in.

6 0

3 years ago

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small cubes with edge lengths of 1/4 inch will be packed into the right rectangular prism shown.( the base is 4 1/2, the width i

ss7ja [257]

General Idea:

We need to find the volume of the small cube given the side length of the small cube as 1/4 inch.

Also we need to find the volume of the right rectangular prism with the given dimension (the height is 4 1/2, the width is 5, and the length is 3 3/4).

To find the number of small cubes that are needed to completely fill the right rectangular prism, we need to divide volume of right rectangular prism by volume of each small cube.

Formula Used:

$Volume \; of \; Cube = a^3 \; \\\{where \; a \; is \; side \; length \; of \; cube\}\\\\Volume \; of \; Right \; Rectangular \; Prism=L \times W \times H\\\{Where \; L \; is \; Length, \; W \; is \; Width, \;and \; H \; is \; Height\}$

Applying the concept:

Volume of Small Cube:

$V_{cube}= (\frac{1}{4} )^3= \frac{1}{64} \; in^3\\\\V_{Prism}= 3 \frac{3}{4} \times 5 \times 4 \frac{1}{2} = \frac{15}{4} \times \frac{5}{1} \times \frac{9}{2} = \frac{675}{8} \\\\Number \; of \; small \; cubes= \frac{V_{Prism}}{V_{Cube}} = \frac{675}{8} \div \frac{1}{64} \\\\Flip \; the \; second \; fraction\; and \; multiply \; with \; the \; first \; fraction\\\\Number \; of \; small \; cubes \;= \frac{675}{8} \times \frac{64}{1} = 5400$

Conclusion:

The number of small cubes with side length as 1/4 inches that are needed to completely fill the right rectangular prism whose height is 4 1/2 inches, width is 5 inches, and length is 3 3/4 inches is <em><u>5400 </u></em>

4 0

3 years ago

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4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

galben [10]

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Rearranging these two equations:

$\\ -0.6*\sigma = 100-\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

$\\ 0.6*\sigma = -100+\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

Summing both equations, we obtain the following equation:

$\\ 3.0*\sigma = 90$

Then

$\\ \sigma = \frac{90}{3.0} = 30$

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

$\\ 2.4*30 = 190-\mu$ (2)

$\\ 2.4*30 - 190 = -\mu$

$\\ -2.4*30 + 190 = \mu$

$\\ \mu = 118$

7 0

3 years ago

How to do improper fractions with 3.8 repeating

vivado [14]

3.88888=3.88888/1=38.8888888

so lets ssay our repeating decimal = a fraction x
x=3.888 is true
multiply by 10
10x=38.8888
now subtract fir frist eqation from second
10x-x=38.8888-3.888888
9x=35
divide both sides by 9
x=35/9
x=3 and 7/9

anser is 3.88=3 and 7/9

3 0

3 years ago

A window washer cleaned 38 windows in 2 hours. At this rate, how many windows did he clean in 7 hours?

andriy [413]

Answer:

Step-by-step explanation:

38/2=19x7=133

6 0

3 years ago

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