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Gre4nikov [31]
2 years ago
11

Solve by multiplying

Mathematics
1 answer:
ANEK [815]2 years ago
6 0
The answer is 70,000
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2, 3, 5, 9, 17...
BARSIC [14]

Answer:

Step-by-step explanation:

i'm sure the answer is C

8 0
3 years ago
Read 2 more answers
Anyone plz help I need it
son4ous [18]

Answer:

27.5 inches

Step-by-step explanation:

(68.1+11.2+x)/3=35.6

68.1+11.2+x=106.8

79.3+x=106.8

x=27.5

4 0
3 years ago
I need help on #5,6, and 7
denpristay [2]
5.
f(3) = 4 \times 3 - 3 = 12 - 3 = 9
6.
f(0) = 4 \times 0 - 3 = 0 - 3 = - 3
7.
f( - 7) = 4 \times ( - 7) - 3 = - 28 - 3 = - 31
8 0
3 years ago
Can you please help me? Keep it a little basic. (no decimals) Thanks for the help!
Delvig [45]
When you are dividing fractions you just flip the second fraction upside down and multiply the two fractions
So the first is
½ x 6/1 = 6/2 = 3

For the second you write them as improper fractions then do the same
So it is
3/2 / 1/8 = 3/2 x 8/1 = 24/2 = 12

So the answer to the first question is 3 and the second is 12
4 0
3 years ago
Read 2 more answers
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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