Find (fxg)(a)?thanks !!

Need Answer Fast Please to solve the Linear Expression: 5(5n+2)+(3-2n)

tatyana61 [14]

Answer:

23n +13

Step-by-step explanation:

5(5n+2)+(3-2n)

Distribute

25n + 10 +3 -2n

Combine like terms

25n -2n + 10+3

23n +13

3 0

3 years ago

Read 2 more answers

Miranda’s financial aid stipulates that her tuition not exceed $1200. If her college charges a $90 registration fee for the term

Arlecino [84]

Answer: the greatest number of courses for which miranda can register is 2

Step-by-step explanation:

Let x represent the number of courses for which Miranda can register.

If her college charges a $90 registration fee for the term plus $475 per course, it means that the total amount that the college charges for x courses would be

475x + 90

Miranda’s financial aid stipulates that her tuition not exceed $1200(it must be at most $1200). This means that

475x + 90 ≤ 1200

475x ≤ 1200 - 90

475x ≤ 1110

x ≤ 1110/475

x ≤ 2.34

3 0

4 years ago

The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 5 cm/s. When

atroni [7]

Answer:

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

Step-by-step explanation:

The area for a rectangle is given by the formula:

$A=w\ell$

Where w is the width and l is the length.

We are given that the length of the rectangle is increasing at a rate of 6 cm/s and that the width is increasing at a rate of 5 cm/s. In other words, dl/dt = 6 and dw/dt = 5.

First, differentiate the equation with respect to t, where w and l are both functions of t:

$\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left[w\ell]$

By the Product Rule:

$\displaystyle \frac{dA}{dt}=\frac{dw}{dt}\ell +\frac{d\ell}{dt}w$

Since we know that dl/dt = 6 and that dw/dt = 5:

$\displaystyle \frac{dA}{dt}=5\ell + 6w$

We want to find the rate at which the area is increasing when the length is 12 cm and the width is 4 cm. Substitute:

$\displaystyle \frac{dA}{dt}=5(12)+6(4)=84\text{ cm}^2\text{/s}$

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

8 0

3 years ago

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

This graph represents which function?

il63 [147K]

Answer:

A

Step-by-step explanation:⁻7+

4 0

4 years ago