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ArbitrLikvidat [17]
3 years ago
6

#16 i

Mathematics
1 answer:
Vlad1618 [11]3 years ago
4 0

The tuition is an illustration of a linear function.

The cost of tuition and fees in the academic year 2023-2024, is $12260

Let the number of academic years after 2014-2015 academic year be x.

So, we have:

\mathbf{(x,y) \to (0,9200)(5,10900)}

A linear function is represented as:

\mathbf{y = mx + b}

Where m represents the slope (i.e. constant rate), and b represents the y-intercept (i.e. the value of y when x = 0)

So, we have:

\mathbf{10900 = 5m + 9200}

Subtract 9200 from both sides

\mathbf{1700 = 5m}

Divide both sides by 5

\mathbf{340 = m}

So, we have:

\mathbf{m =340 }

The function becomes

\mathbf{y = mx + b}

\mathbf{y = 340x + 9200}

In the academic year 2023-2024, x = 9.

So, we have:

\mathbf{y = 340x + 9200}

\mathbf{y = 340 \times 9 + 9200}

\mathbf{y = 3060 + 9200}

\mathbf{y = 12260}

Hence, the cost of tuition and fees is $12260

Read more about linear functions a:

brainly.com/question/21107621

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Answer:

The correct answer is: Option D) 5

Step-by-step explanation:

Given equation is:

3(x + 2) + x = 2x + 16

In order to find that which values of x makes the equation true, we have to put each value of x in the equation. When both sides of equations will be equal, that value of x will be true for the equation.

Putting x = 2

3(2 + 2) + 2 = 2(2) + 16\\3(4)+2 = 4+16\\12+2 = 20\\14 \neq 20

Putting x = 3

3(3 + 2) + 3 = 2(3) + 16\\3(5)+3 = 6+16\\17+3 = 22\\20 \neq 22

Putting x=4

3(4+2)+4 = 2(4)+16\\3(6)+4 = 8+16\\18+4 = 24\\22 \neq 24

Putting x = 5

3(5+2)+5 = 2(5)+16\\3(7)+5 = 10+16\\21+5 = 26\\26 = 26

The equation is true for x = 5

Hence,

The correct answer is: Option D) 5

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3 years ago
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Sever21 [200]

Answer:

7512 hours

Step-by-step explanation:

There are 8760 hours in a year.

There are 52 hours on Thursdays and Saturdays in a year.

The number of hours for Thurdays and Saturdays in a year is 1248.

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6 0
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Rama09 [41]

Answer:

a) E(X) = 6.45

b) E(X^{2} )= 57.25

c) V(X) = 15.648

d) E(3X + 2) = 21.35

e) E(3X^{2} +2) = 173.75

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) E(X) = \sum xP(x)

E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45

b)

E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\  E(X^{2} )= 57.25

c)

V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648

d)

E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35

e)

E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75

f)

V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832

g)

E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45

h)

V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648

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