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3 years ago

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Is this one question correct

1 answer:

bazaltina [42]3 years ago

7 0

Answer: 2/3ft or C

Step-by-step explanation: Solve for b in the area of a triangle equation, A=1/2bh. Multiply 1/2 by 2 on both sides to cancel it: A(2)=bh. Divide both sides by h: $\frac{A(2)}{h}$ =b

A = 2/5

h = 6/5

b = ?

Next, we can plug the numbers into this equation: $\frac{2/5(2)}{6/5}$ =b

Multiply 2 and 2/5: 4/5 / 6/5

Take the reciprocal of 6/5: 4/5 x 5/6 = $\frac{20}{30}$

$\frac{20}{30}$ ÷ $\frac{10}{10}$ = $\frac{2}{3}$

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If you need to put 28.93 gallons of ice cream into just cups, how many cups can you make out of these? Round to a whole number.

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Answer:

16 cups = 1 gallon

16( 28.93) = 462.88

Therefore, there are 462 cups of ice cream.

Step-by-step explanation:

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4 years ago

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

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Answer:

You can save 0.40 if you choose to buy the 16-ounce bottles instead of 10-ounce bottles.

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2 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

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3 years ago

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Step-by-step explanation:

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