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Mashcka [7]
3 years ago
15

Determine the slope and point on the line for each equation C. y+ 4 = %(% - 3)

Mathematics
1 answer:
Temka [501]3 years ago
5 0
Bbbbhbjjjjioiiu and his friends were the
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PLZ I NEED HELP WITH THIS QUESTIOn
lesya692 [45]

Answer:

12 units

Step-by-step explanation:

there are 3 sqaures counted from (0,0) to (3,0)

so do 3 + 3 + 3+ 3 and get 12 units

if brainiest is earned its greatly Appreciated

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
F(x) = x2<br> What is f(x) + f(x) + f(x)?
Yuliya22 [10]

Answer:

if you meant x squared the answer is 3x^2

Step-by-step explanation:

If f(x) is equal to 2x you can substitute x squared for f(x) and that would be x squared+x squared+x squared which is 3x^2

6 0
3 years ago
Read 2 more answers
The LCM of 12 and 6 is 6. True or False
elena-s [515]

Answer:

false but Least common multiple (LCM) of 6 and 12 is 12.

7 0
3 years ago
Read 2 more answers
A bag contains an unknown number of marbles. You know that P(red) = 1 and P(green) = 1. What can you conclude about how many mar
Nataly [62]

The number of marbles in the bag illustrates probability

The conclusion about the number of marbles in the bag is that the bag can only contain one type of marble

<h3>The number of marbles in the bag?</h3>

The probabilities are given as:

P(Red) = 1

P(Green) = 1

For a distribution, the sum of all probabilities must equal to 1.

This means that:

P(Red) + P(Green) = 1

So, we have:

1 + 1 = 1

This gives

2 = 1

The above equation is false because 2 does not equal to 1.

So, the conclusion about the number of marbles in the bag is that the bag can only contain one type of marble i.e. either red marbles are in the bag or blue marbles are in the bag

Read more about probability at:

brainly.com/question/251701

5 0
2 years ago
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