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3 years ago

8

The Area of a rectangular patio is 5 5/8 square yards,and it's length is 1 1/2 yards.What is the patio's width,in yards?

1 answer:

fgiga [73]3 years ago

8 0

Take both mixed numbers (a whole number plus a fraction) and turn them into improper fractions ( a numerator (top number) bigger than the denominator (bottom number). To do this, multiply the denominator by the whole number, and add the result to the numerator.
5 5/8 will become 45/8
1 1/2 will become 3/2.
Area is length* width, so width would be Area/ length. So 45/8 is the area and 3/2 is the length. Now you have (45/8)/(3/2). To divide fractions, take the second fraction and flip the numerator and denominator, forming the reciprocal (2/3). Now multiply 45/8 and 2/3
The result is 90/24, and simplifying would get 15/4, or 3 3/4. This is the width(or length)

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Read 2 more answers

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umka2103 [35]

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The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of <em>X</em> as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

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