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Zina [86]
2 years ago
7

a rectangle has a perimeter of 3o inches and a width of 6 inches what are the dimensions of a similar rectangle with a perimeter

of 20 inches?
Mathematics
1 answer:
xz_007 [3.2K]2 years ago
3 0

Answer:

the width would be 4 inches and the length would be 6 inches

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A rectangle has a height of 6k3 and a width of 2k2 + 4k +5.
Trava [24]

Answer: 12k^5+24k^4+30k^3

Step-by-step explanation:

Given : The height of the rectangle = 6k^3

The width of the rectangle = 2k^2+4k+5

Formula : Area = height x width

Therefore , the area of triangle in terms of polynomial will be :

6k^3\times( 2k^2+4k+5)\\\\= 6k^3(2k^2)+6k^3(4k)+6k^3(5)\ \ [\text{Using Distributive property}]\\\\=12k^{3+2}+24k^{3+1}+30k^3\ \ [\text{Using exponents rule}:\ a^n\times a^m=a^{n+m}]\\\\=12k^5+24k^4+30k^3

Hence, the area of the entire rectangle =12k^5+24k^4+30k^3

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3 years ago
Is 1/4 more than 7/3
IrinaVladis [17]

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A food company sells salmon to various customers. The mean weight of the salmon is 37 lb with a standard deviation of 2 lbs. The
brilliants [131]

Answer:

The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.

Step-by-step explanation:

The mean sample of the sum of n random variables is

\overline{X} = \frac{X_1+X_2+...+X_n}{n}

If X_1, ..., X_n are indentically distributed and independent, like in the situation of the problem, then the variance of X_1 + .... + X_n will be the sum of the variances, in other words, it will be n times the variance of X_1 .

However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus \overkine{X} = \frac{V(X_1)}{n} and as a result, the standard deviation of \overline{X} is the standard deviation of X_1 divided by \sqrt{n} .

Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:

  • Restaurants: We have boxes with 9 salmon each, so it will be \frac{2}{\sqrt{9}} = \frac{2}{3}
  • Grocery stores: Each carton has 49 salmon, thus the standard deviation is \frac{2}{\sqrt{49}} = \frac{2}{7}
  • Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is \frac{2}{\sqrt{64}} = \frac{1}{4}

We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.

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3 years ago
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