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oee [108]
3 years ago
11

The volume of a cube shaped shipping container is 2744 in.³ find the edge length of the shipping container

Mathematics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

The edge length of the shipping container is 14 in.

Step-by-step explanation:

The volume enclosed by a cube  is the number of cubic units that will exactly fill a cube.

To find the volume of a cube  we need to recall that a cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself twice. Or as a formula:

                                                        V=s^3

where,  <em>s</em> is the length of any edge of the cube.  

To find the edge length of the shipping container we use the fact that the volume of a cube shaped shipping container is 2744 in³ and the above formula.

2744=s^3\\\\s^3=2744\\\\s=\sqrt[3]{2744}\\\\s = 14\:in

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A random sample of 9 wheels of cheese yielded the following weights in pounds has a sample mean of 20.90 and a sample variance o
Ronch [10]

Answer:

2.002 \leq \sigma^2 \leq 11.365

Step-by-step explanation:

1) Data given and notation

s represent the sample standard deviation

s^2 represent the sample variance

n=9 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^9 (x_i -\bar x)^2}{n-1}}&#10;The sample variance given was [tex]s^2=3.45

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,8)" "=CHISQ.INV(0.95,8)". so for this case the critical values are:

\chi^2_{\alpha/2}=15.507

\chi^2_{1- \alpha/2}=2.732

And replacing into the formula for the interval we got:

\frac{(9)(3.45)}{15.507} \leq \sigma \frac{(9)(3.45)}{2.732}

2.002 \leq \sigma^2 \leq 11.365

7 0
3 years ago
William is going to invest in an account paying an interest rate of 5.9% compounded
Valentin [98]

Answer:

\$400.92  

Step-by-step explanation:

we know that

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

t=19\ years\\ A=\$1,230\\ r=5.9\%=5.9/100=0.059  

substitute in the formula above

1,230=P(e)^{0.059*19}  

solve for P

1,230=P(e)^{1.121}  

P=1,230/(e)^{1.121}

P=\$400.92  

3 0
4 years ago
Do you need to ask a question, but don't have points.....
horsena [70]
Yes, felt that. thanks and have a lovely day!:D
8 0
2 years ago
Read 2 more answers
Abe real estate is developing 15 solar homes per states. If the cost of each home is estimated at 80,000 what’s the projected co
Soloha48 [4]

Answer:

the Total cost= 1,200,000*number of states

Step-by-step explanation:

Given the  cost of each home is $80,000

The number of homes per state are 15

The Total cost of the project is = (Total number of houses per state) * (cost of each house) * (the number of states).

Therefore the total cost =15*80000*Number of states.=1,200,000*number of states

3 0
3 years ago
Can someone please help?!
attashe74 [19]

                   .                                                                    

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. .                                                              

                                              .       .

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3 years ago
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