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SashulF [63]
2 years ago
10

A jar contains 4 blue, 11 pink, 5 red, and 6 orange jellybeans. Once is selected, replaced, then another is selected. Find P(blu

e, then red) really need this correct
Mathematics
1 answer:
faltersainse [42]2 years ago
4 0

Answer:

P(blue then red)=5/169

Step-by-step explanation:

the probability of picking the blue is 4/26 it's replaced so the total jellybeans doesn't change therefore the sample space doesn't change. the probability of picking a red is 5/26

to find the probability of picking them in that order, you multiply the two together

(4/26)x(5/26)=20/676

reducing that we get:

P(blue then red)=5/169

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Sarah's favorite candy consists of sour patch kids and M&Ms. She has 17 total packages of sour patch kids and M&Ms. She
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Sarah has 11 packages of sour patch kids and 6 packages of M&Ms

Step-by-step explanation:

11 + 6 = 17, and 11 is five more than 6, fulfilling the needs of the problem

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Solve <br> y=1/4x+5 y=2x-9
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Step-by-step explanation:


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3 years ago
Bob discovers a box with 30 identical, used, rechargeable batteries in his attic.
monitta

Answer:

a) All of them are out of charge = 9.31x10⁻¹⁰

b) 20% of them are out of charge = 5.529x10⁻⁴

Step-by-step explanation:

This problem can be modeled as a binomial distribution since

There are n repeated trials and all of them are independent of each other.

There are only two possibilities: battery is out of charge and battery is not out of charge.

The probability of success does not change with trial to trial.

Since it is given that it is equally likely for the battery to be out of charge or not out of charge so probability of success is 50% or 0.50

P = 0.50

1 - P = 0.50

a) All of them are out of charge?

Probability = nCx * P^x * (1 - P)^n-x

Probability = ₃₀C₃₀(0.50)³⁰(0.50)⁰

Probability = 9.31x10⁻¹⁰

b) 20% of them are out of charge?

0.20*30 = 6 batteries are out of charge

Probability =₃₀C₆(0.50)²⁴(0.50)⁶

Probability = 5.529x10⁻⁴

5 0
3 years ago
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Darya [45]

Answer:  D) csc (2π/3)

<u>Step-by-step explanation:</u>

2π/3 is in Quadrant II

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y-values are: sin & csc

tan = y/x so will be negative in Quadrant II

A) sec = -

B) tan = -

C) cos = -

D) csc = +     this works!

7 0
3 years ago
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