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Marta_Voda [28]
2 years ago
5

Colin and amil Each built boxes with a volume of 8 ft.³ Collins box is a cube almils box is a rectangular prism the length of am

ils box is twice the length of Collins box all of the side measures are whole numbers what could be the length width and height of amils box
Mathematics
1 answer:
SOVA2 [1]2 years ago
3 0

Answer:

4' x 2' x 1'

Step-by-step explanation:

Collins' cube has a volume of that is the length of any side, x, cubed:  Vol = x^3.  Since his box has 8^3, we can say that x = 2.  <u>[2^3 = 8]</u>

Amil's box has one side that is 2x.  That side would be 2*2 = 4 feet.  His volume is also 8 ft^3.  Amil's box also has a volume of 8 ft^3.

His box dimensions are therefore:  (4)(X)(Y) = 8 ft^3 , where X and Y are whole-number dimensions for the other 2 dimensions of his box.

(4)(X)(Y) = 8 ft^3

      X*Y = 2

The only combination of whole numbers for which this this would work is 1 and 2.

Amil's box is 4' x 2' x 1' or 8 ft^3

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Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3
aliina [53]

Answer:

C) more than the absolute value of the critical value, we can conclude that the average monthly rate for an assisted-living facility is not equal to $3,300.

Step-by-step explanation:

Hello!

Interest hypothesizes is " the average monthly rate for one-bedroom assisted-living facility is equal to $3300" symbolically: μ = 3300

The study variable is:

X: Monthly rate for a one-bedroom assisted-living facility.

Since there is no information about the distribution of the variable, to be able to study the population mean, I'll assume that the variable has a normal distribution.

The hypothesis is:

H₀: μ = 3300

H₁: μ ≠ 3300

α: 0.05

The statistic to use, considering that there is no known population information and the sample size, is a Student t:

t= <u> X[bar] - μ </u> ~t_{n-1}

       S/√n

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X[bar]= $3690

S= $530

Using the sample data, calculate the statistic value:

t= <u> 3690 - 3300 </u> = 2.549

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The rejection region for this test is two-tailed, with critical values:

t_{n-1; \alpha /2} = t_{11; 0.025} = -2.201

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The decision rule is:

Reject the null hypothesis if t ≤ -2.201 or if t ≥ 2.201

Not reject the null hypothesis if -2.201 < t < 2.201

Since the calculated value (2.549) is greater than the right critical value (2.201) the decision is to reject the null hypothesis.

With a signification level of 5%, there is enough evidence to reject the null hypothesis. This means that the population mean of the monthly rate for a one-bedroom assisted living facility is different from $3300.

I hope it helps!

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