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1 year ago

Example 1:

Mathematics

1 answer:

Charra [1.4K]1 year ago

7 0

Answer: f(4) = 8, f(-5) = -19, f(2-x) = -3x+2

Step-by-step explanation:

Let's start by analyzing what we are given:

f(x)=3x-4

In this equation, f is the name of the function.

The number in the parentheses represents x.

The 3x-4 tells us what the equation does. In this instance, we multiply x by 3 and then subtract 4.

Let's start with (a).

f(4)

Since we have established that the number in the parentheses is x, we can plug 4 in for x into the equation f. We must abide by the order of operations. Therefore, we can obtain the following:

f(4) = 3(4) -4

f(4) = 12 -4

f(4) = 8

We can continue this for part (b).

f(-5) = 3(-5) -4

f(-5) = -15-4

f(-5) = -19

Part (c) is slightly different, but applies the same principles we established previously. On this part, we must recall the distributive property. The 3 multiples the 2 and the x. We must also remember to combine like terms.

f(2-x) = 3(2-x) -4

f(2-x) = 6 - 3x -4

f(2-x) = -3x +2

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(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal( $\mu=237,\sigma^{2} =54^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean family spending = $237

$\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

= 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

= 1 - 0.8749 = 0.1251

P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

= 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101

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