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jeka94
3 years ago
5

A rectangular box is to have a square base and a volume of 70 ft3. The material for the base costs 38¢/ft2, the material for the

sides costs 10¢/ft2, and the material for the top costs 27¢/ft2. Letting x denote the length of one side of the base, find a function in the variable x giving the cost (in dollars) of constructing the box.
Mathematics
1 answer:
lakkis [162]3 years ago
3 0
x-length\ of\ one\ side\ of\ base\\y-high\ of\ box\\x^2-area\ of\ base\\x^2y-volume\\\\x^2y=70\\y=\frac{70}{x^2}\ \ \ \ \ \ \ \ and\ x\neq0\\\\38x^2+4\cdot10xy+27x^2\\65x^2+40xy\ in\ cents\\\\0.65x^2+0.4xy\ in\ dollars
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I NEED THE ANSWER(ASAP)
Ivan

Answer:

9.25

Step-by-step explanation:

For the mean, you have to add all and divide it by the numbers.

7 + 12 + 8.50 + 10 + 7 + 8 + 10.50 + 11 = 74

74 divided by 8 = 9.25

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2 years ago
As a member of a music club you can order CDs for $14.99 each the music club also charges $4.99 for each shipment the expression
alex41 [277]
It would cost 49.96 to order three Cd's. 
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Hope this helps

4 0
3 years ago
What value digit in 2,417 is ten times as value of the digit in 2,147
3241004551 [841]

Answer:

4 in 2,417

4 in 2,147

Step-by-step explanation:

the place value of 4 in 2,417 is HUNDREDS

while the place value of 4 in 2,147 is TENS

Therefore the place value of 4 in 2,417 is ten times the value of 4 in 2,147

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3 years ago
Find the expected value and the variance of the number of times one must throw a die until the outcome 1 has occurred 5 times.
Kazeer [188]

Answer: E(X) = 30; Var[X] = 180

Step-by-step explanation: This is a <u>Bernoulli</u> <u>Experiment</u>, i.e., the experiment is repeated a fixed number of times, the trials are independents, the only two outcomes are "success" or "failure" and the probability of success remains the same, So, to calculate <em><u>Expected</u></em> <em><u>Value</u></em>, which is the mean, in these conditions:

E(X)=\frac{r}{p}

r is number of times it is repeated

p is probability it happens

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E(X)=\frac{5}{1/6}

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<u>Variance</u> is given by:

Var[X]=\frac{r(1-p)}{p^{2}}

Var[X]=\frac{5(1-1/6)}{(1/6)^{2}}

Var[X]=5.\frac{5}{6}.6^{2}

Var[X] = 180

Expected Value and Variance of the number of times one must throw a die until 1 happens 5 times are 30 and 180, respectively.

4 0
3 years ago
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Triss [41]
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5 0
3 years ago
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