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Phoenix [80]
3 years ago
8

Cal's Coffee prints prize coupons under the rims of 20% of it's paper cups . If you buy ten cups of coffee:

Mathematics
1 answer:
Softa [21]3 years ago
4 0

For the answer to the question above the probability of exactly x successes is
P(X=x)=b(x;n,p) = (nCx)(p^x)((1-p)^(n-x))
where nCx is number of combinations of n things taken x at a time, and "^" means exponentiation.
n = 10  
p = 0.2  

P(X=7) = 10C7*(0.2^7)*(0.8^3) = 0.00079 
P(X=8) = 10C8*(0.2^8)*(0.8^2) = 0.00007 
P(X=9) = 10C9*(0.2^9)*(0.8^1) = 0.00000 
P(X=10) = 10C10*(0.2^10)*(0.8^0) = 0.00000 

.00079 + .00007 + 0 + 0 = .00086 = .0009 rounded 
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Step-by-step explanation:

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A square with an area of 64 in2 is rotated to form a cylinder. What is the radius of the cylinder?
romanna [79]
The area of the square is calculated through the equation,

                               Area = e²

where e is the measure of sides.
 
From the given,

                              64 = e²
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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
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