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erma4kov [3.2K]

3 years ago

8

The volume of a rectangular prism with a length of x meters, a width of x − 1 meters, and a height of x + 11 meters is no more t

han 180 cubic meters. What are the possible values of the length?

2 answers:

Rashid [163]3 years ago

8 0

Answer:

Length of the rectangular prism = 4 meters

But other possible values = (-5meters or - 9 meters)

Step-by-step explanation:

The volume of a rectangular prism = Length × Width × Height

From the question above,

Length = x meters

Width = x - 1 meters

Height = x + 11 meters

Volume of the Rectangular prism = 180 cubic meters

Hence,

(x) × (x - 1) × (x + 11) = 180

We expand the brackets

(x)(x - 1) (x + 11) = 180

x² - x(x + 11) = 180

x² (x + 11) - x(x + 11) = 180

x³ + 11x² - x² + 11x =180

x³ +10x² - 11x = 180

x³ + 10x² - 11x -180 = 0

The above is a polynomial

We solve this polynomial to find x

x³ + 10x² - 11x -180 = 0

(x - 4)(x + 5) (x + 9) = 0

x - 4 = 0

x = 4

x + 5 = 0

x = -5

x + 9 = 0

x = -9

We are asked to find the various values for the length hence,

From the above question, we are told that

Length = x meters

Therefore, the length of this rectangular prism = 4 meters or -5 meters or -9 meters.

Lena [83]3 years ago

7 0

Answer:

(1, 4)

Step-by-step explanation:

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Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

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therefore for a Binomial distribution where

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since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] = 1

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Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

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Answer:

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Given that, each successive row contains two fewer seats than the preceding row.

Formula:

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l= last term

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$T_n=a+(n-1)d$

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Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

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$\Rightarrow t = \frac{n+1}{2}$

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$=\frac{n^2+2n+1}{4}$

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Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

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⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

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