AB has endpoints at (-4,5) and (4,-3) using the midpoint formula what is the midpoint segment

Need help ASAP and will give brainliest if right!

Luba_88 [7]

Answer: The answer is converges r=1/7

4 0

3 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

2 years ago

An isotope decays at a rate of -0.0049% per year. Find the half life. (show the work)

cricket20 [7]

Answer:

14,146 years

Step-by-step explanation:

The annual multiplier is (1 -0.0049%) = 0.999951. We want to find the number of times this needs to be multiplied to make the product 1/2.

0.5 = 0.999951^t

log(0.5) = t·log(0.999951)

log(0.5)/log(0.999951) = t ≈ 14,145.514

The half-life is about 14,146 years.

7 0

3 years ago

Write an equation of the circle with center (6, 2) and radius 4.

luda_lava [24]

Answer:

$(x-6)^2+(y-2)^2=16$ .

Step-by-step explanation:

$(x-h)^2+(y-k)^2=r^2$ is the equation for a circle with center (h,k) and radius r.

You are given center (6,2) and radius 4.

So we will replace h with 6 and k with 2 and r with 4.

This gives us:

$(x-6)^2+(y-2)^2=4^2$

Simplify:

$(x-6)^2+(y-2)^2=16$ .

8 0

2 years ago

Read 2 more answers

A football team recieved 3 penalties for 15 yeards each 1 for 5 yards and 1 for half the distance to tge goal from the 34-yard l

Oksanka [162]

He w has 59 penalties

6 0

2 years ago