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Ksju [112]
2 years ago
13

Find the area of the triangle below.

Mathematics
1 answer:
Rom4ik [11]2 years ago
6 0

Step-by-step explanation:

the area of a triangle is

baseline × height / 2

in our case we have a baseline with its associated height right there. and so the area is

17 × 6 / 2 = 17 × 3 = 51 ft²

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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
In Ms. Craig’s class, 75% of the students are boys. There are 18 boys in the class. What is the total number of students in Ms.
topjm [15]

Answer:

24

Step-by-step explanation:

75% of x = 18

75/100 * x = 18 (x =  total number of students)

solving

x = 24

total number of students = 24

Checking:

75/100 * 24 = 18

5 0
3 years ago
Bob makes 4% commission on his sales as a real estate agent. If Bob sells a house for $249,000, how much money does he get to ke
Lana71 [14]
He gets to keep $9,960 of the sale
4 0
3 years ago
Write an equation to find the amount of interest $450,000 earns in 1 year at a simple interest rate of 3.5%.
telo118 [61]

Answer:

The required equation for the amount of Simple Internet = \frac{450,000 \times 1 \times 3.5}{100}

Step-by-step explanation:

Here , Principal amount = $450,000

Time  = 1 year

Rate of Interest = 3.5%

Now,SIMPLE INTEREST  = \frac{P \times R \times T}{100}

So, here SI  = \frac{450,000 \times 1 \times 3.5}{100}

Hence, the required equation for the amount of Simple Internet = \frac{450,000 \times 1 \times 3.5}{100}

7 0
3 years ago
Read 2 more answers
The heights of women aged 20 to 29 follow approximately the N(64, 2.82) distribution. Men the same age have heights distributed
ivann1987 [24]

Answer:96.97

Step-by-step explanation:

Given

mean height of women \mu _w=64

Standard deviation of height \sigma _w=2.82

mean height of men \mu _=69.3

Standard deviation of height \sigma _m=2.71

P\left ( X

P\left ( \frac{x-\mu _w}{\sigma _w}\right )

P\left ( Z

i.e. 0.9697=96.97 %

4 0
3 years ago
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