11.) Solve the equation.

Simplify 8x^2-x-12x+2x2

balu736 [363]

Answer:

Simplify: 8 2 − 1 3 + 2 2

Step-by-step explanation:

Solution: 1 0 2 − 1 3

3 0

3 years ago

You save for retirement over 30 years by investing $850/month in a stock account that yields 10%. You invest $350/month in a bon

sergejj [24]

Answer:

$13,287.70

Step-by-step explanation:

The future value of the stock account is computed as the sum of a geometric series. This computation assumes that the annual yield is compounded monthly.

FV = p((1+r/12)^(12n) -1)/(r/12)

For the stock account, p=850, r=0.10, n=30, so the future values is ...

FV = 850((1+.10/12)^360-1)/(.10/12)) = 1,921,414.74

For the bond account, p=350, r=.06, n=30, so the future value is ...

FV = 350((1+.06/12)^360 -1)/(.06/12) = 351,580.26

The combined account value at the end of 30 years is ...

$1,921,414.74 + 351,580.26 = $2,272,995.00

_____

The monthly payment that can be made over a 25 year period is given by the amortization formula.

A = P(r/12)/(1 -(1 +r/12)^(-12n))

= $2,272,995.00(.05/12)/(1 -(1+.05/12)^-300) = $13,287.70

You can withdraw $13,287.70 each month assuming a 25-year withdrawal period.

6 0

4 years ago

What is the volume, in cubic centimeters, of a rectangular prism with a height of 17 centimeters, a width of 17 centimeters, and

leonid [27]

The answer is V 3179 cm^3

5 0

3 years ago

Read 2 more answers

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time).

NeX [460]

Answer:

(a) PRIVATE COLLEGES

Sample mean is $42.5 thousand

Sample standard deviation is $6.65 thousand

PUBLIC COLLEGES

Sample mean is $22.3 thousand

Sample standard deviation is $4.34 thousand

(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)

(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand

Step-by-step explanation:

(a) PRIVATE COLLEGES

Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand

Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand

PUBLIC COLLEGES

Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand

Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand

(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.

This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges

(c) Confidence Interval = Mean + or - Margin of error (E)

E = t×sd/√n

Mean = $42.5 - $22.3 = $20.2 thousand

sd = $6.65 - $4.34 = $2.31 thousand

n = 10+12 = 22

degree of freedom = 22-2 = 20

t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086

E = 2.086×$2.31/√22 = $1.0 thousand

Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand

Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand

95% confidence interval is $19.2 thousand to $21.2 thousand

6 0

3 years ago

Which of the following place value tables shows 40,833?

jolli1 [7]

40,833 has...

4 Ten Thousands

8 Hundreds

3 Tens

3 Ones

6 0

4 years ago