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Kazeer [188]
3 years ago
14

A store charges their customers based on the number of items that they buy. Jane bought four items and was charged $15 and Georg

e bought twelve items and was charged $39. Write a function to model the cost based on the number of items you buy. Then determine how much you bought if you were charged $33.
A) f(x) = 3x + 3; f(10) = 33
B) f(x) = 3x + 3; f(33) = 102
C) f(x) =1/3 + 3; f(90) = 33
D) f(x) = 1/3x + 3; f(33) = 14
Mathematics
1 answer:
mojhsa [17]3 years ago
7 0

Answer:

A) f(x) = 3x + 3; f(10) = 33

Step-by-step explanation:

39 = 3x + 3 \\ \\ 36 = 3x \\ \\ 12 = x

15 = 3x + 3 \\ \\ 12 = 3x \\ \\ 4 = x

*33 = 3x + 3 \\ \\ 30 = 3x \\ \\ 10 = x

** From what the problem said, we know that we selected the correct equation.

I am joyous to assist you anytime.

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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. The
zubka84 [21]

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

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

P(X = 0) =(0.8)^{7

P(X = 0) =0.2097152

If it rains on any one day; then X = 100

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

P(X = 100) =  _7C_1 * 0.2^1 * (1-0.2)^{7-1

P(X = 0) =7 * 0.2* (1-0.2)^{6

P(X = 100) =7* 0.2* (0.8)^{6

P(X = 100) =0.3670016

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

P(X = 200) =  21 * 0.2^2 * (0.8)^{5

P(X = 200) = 0.2752512

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

(E [X]) = \sum x * p\\ \\  (E [X]) =  0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55

NOW;

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

the variance of the refund payment to the couple = 9463.394

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