10 points help please help me no understand math

The proportions of blood phenotypes, A, B, AB, and O, in the population of all Caucasians in the United States are approximately

Vanyuwa [196]

Answer:

a. S = {A,B,AB,O}

b. P(A) = 0.41, P(B) = 0.1, P(AB) = 0.04, P(O) = 0.45

c. 0.45 = 45% probability that the person chosen at random has either type A or type AB blood

Step-by-step explanation:

a List the sample space for this experiment.

All possible blood types(A, B, AB or O). So

S = {A,B,AB,O}

b Make use of the information given above to assign probabilities to each of the simple events.

The proportions of blood phenotypes, A, B, AB, and O, in the population of all Caucasians in the United States are approximately .41, .10, .04, and .45, respectively.

So

P(A) = 0.41, P(B) = 0.1, P(AB) = 0.04, P(O) = 0.45

c What is the probability that the person chosen at random has either type A or type AB blood

P(A) + P(AB) = 0.41 + 0.04 = 0.45

0.45 = 45% probability that the person chosen at random has either type A or type AB blood

6 0

2 years ago

Please solve this problem, I have no idea with it

lord [1]

Answer:

(x-3)(x-6)

Step-by-step explanation:

Fill in g into f and simplify:

(x-2)²-5(x-2)+4 =

x² - 4x + 4 - 5x + 10 + 4 =

x² - 9x + 18 =

(x-3)(x-6)

4 0

3 years ago

A student received a 75% on an exam. If there were 20 questions on the exam, which ratio would show the number of questions answ

Agata [3.3K]

75%=¾
¾*20=15

He answered 15 questions correctly.

6 0

3 years ago

Read 2 more answers

Your grandfather placed $5,000 in a trust fund for you. In 12 years, the fund will be

PolarNik [594]

Answer:

fv/pv=(1+r)^n

10,000/5000=(1+r%)^12

2=(1+r%)^12

r= 5.95

4 0

2 years ago

Read 2 more answers

Pls solve the simultaneous equation in the attachment.

siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> equation B

Multiply equation A by 10 both sides to remove the fractions

$5x-2y=10$ ----> equation C

isolate the variable y in equation B

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

$5x-2y=10$ ----> equation C

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

Multiply by 3 both sides

$15x-2x-48=30$

$15x-2x=48+30$

Combine like terms

$13x=78$

$x=6$

<em>Find the value of y</em>

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

The solution is the ordered pair (6,10)

Part b) we have

$xy=21$ ---> equation A

$x+5y=22$ ----> equation B

isolate the variable x in the equation B

$x=22-5y$ ----> equation C

substitute equation C in equation A

$(22-5y)y=21$

solve for y

$22y-5y^2=21$

$5y^2-22y+21=0$

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

$x=22-5(1.4)=15$

For y=3

$x=22-5(3)=7$

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0

3 years ago