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BARSIC [14]
3 years ago
13

The volume of a rectangular prism is given by the formula V = lwh, where l is the length of the prism, w is the width, and h is

the height. Suppose a box in the shape of a rectangular prism has length (2a + 11), width (5a – 12), and height (a + 6). Which expression represents the volume of the box?
Mathematics
2 answers:
DaniilM [7]3 years ago
6 0

Answer:

Volume = 10a³ + 91a² + 54a - 792

Step-by-step explanation:

In the absence of answer choices, let's find the expression for the volume.

Given: Volume = length×width×height

V = lwh

length =(2a + 11)

width =(5a – 12)

height= (a + 6)

V =  (2a + 11)(5a – 12) (a + 6)

Expand the first two brackets using distributive property

V = (10a² -24a +55a - 132)(a + 6)

Collect like terms

V = (10a² + 31a -132)(a + 6)

Expand the two brackets using distributive property

V = 10a³ + 31a² - 132a + 60a² + 186a - 792

Collect like terms

V = 10a³ + 91a² + 54a - 792

The expression that represents the volume of the box = 10a³ + 91a² + 54a - 792

astra-53 [7]3 years ago
3 0

Answer:

Volume = 10a³ + 91a² + 54a - 792

Step-by-step explanation:

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2 years ago
Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
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Answer: Part a) P(a)=\frac{1}{\binom{r+w}{r}}

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Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = P(E)=\frac{Favourablecases}{TotalCases}

For part a)

Probability that a red ball is drawn in first attempt = P(E_{1})=\frac{r}{r+w}

Probability that a red ball is drawn in second attempt=P(E_{2})=\frac{r-1}{r+w-1}

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Generalising this result

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Part b)

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1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at (r+1)^{th} draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals \binom{r}{r-1}

Thus the probability becomes P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}

Thus required probability of case b becomes P(E)+ P(E_{i})

= P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\

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