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2 years ago

Which expressions are equivalent to the expression below?

Mathematics

1 answer:

Dmitriy789 [7]2 years ago

8 0

Given :-

-1/3 - ( -4 + 1/6 )

To Find :-

The expression that is equivalent to one of the choices given .

Solution :-

As we know that ,

→ (-) × (-) = (+)

→ (-) × (+) = (-)

→ (+) × (-) = (-)

→ (+) × (+) = (+)

On using these open the brackets ,

→ -1/3 - 1( -4 + 1/6 )

→ -1/3 - 1(-4) + -1(+1/6)

→ -1/3 (-)(-)(1* 4) (+)(-) (1*1/6)

On using now above stated rules ,

→ -1/3 +4 -1/6

Somewhat rearrange ,

→ 4 -1/3 -1/6

Take (-) as common,

→ 4 - (1/3 +1/6)

Hence Option (d) & (f) are correct .

I hope this helps.

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Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

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$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

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if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

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if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

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the variance of the refund payment to the couple = 28002.944 - 18539.55

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