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faust18 [17]
2 years ago
12

I can’t figure this out. Please help

Mathematics
2 answers:
patriot [66]2 years ago
7 0

Answer:

itjust algebra

Step-by-step explanation:

bulgar [2K]2 years ago
6 0

Answer:

\frac{3^{6} }{3^{4} }   or 3²

Step-by-step explanation:

Count the 3 and put it in the exhibitor for the numerator and the denominator of the fraction. You have \frac{3^{6} }{3^{4} } then put them in a line 3^{6}  3^{-4} = 3^{6-4} = 3^{2}

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They are increasing at the same rate.

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One round of golf is 18. A hole in one happens once in every 3708 rounds of golf. How many holes of golf are played for every ho
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the half-life of isotope X is 2.0 years. How many years would it take for a 4.0mg sample of X to decay and have only 0.50 mg of
shepuryov [24]

After 2.0 years, the 4.0 mg sample decays to 2.0 mg.

After another 2.0 years (total 4.0 years), the 2.0 mg sample decays to 1.0 mg.

And after another 2.0 years (total 6.0 years), the 1.0 mg sample decays to 0.50 mg.

So it takes 6.0 years for 4.0 mg to decay to 0.50 mg.

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2 years ago
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
2 years ago
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