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3 years ago

20 POINTS! Find P(4). I don't know or understand how to do this.

Mathematics

1 answer:

Tems11 [23]3 years ago

3 0

Its a 1 out of 8 chance to get 4 or a 0.125 chance in landing on it

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1. Find the measurement of angle J.<br><br> 2. (5x - 13) (3x + 17)

Alex Ar [27]

Answer:

Q1. x= 18, y=59

Q2. m∠J= 56°

Step-by-step explanation:

Q1. (3x +5)°= y° (base ∠s of isos. △)

y= 3x +5 -----(1)

(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)

3x +5 +y +4x -10= 180

7x +y -5= 180

7x +y= 180 +5

7x +y= 185 -----(2)

Substitute (1) into (2):

7x +3x +5= 185

10x= 185 -5

10x= 180

x= 180 ÷10

x= 18

Substitute x= 18 into (1):

y= 3(18) +5

y= 59

Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)

5x -13= 3x +17

5x -3x= 17 +13

2x= 30

x= 30 ÷2

x= 15

∠LKJ

= 3(15) +17

= 62°

∠KLJ= 62° (base ∠s of isos. △)

m∠J

= 180° -62° -62° (∠ sum of △JKL)

= 56°

8 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$