Hey, can anyone solve this real quick? (It's pretty easy but my brain has melted today.)

Mathematics

2 answers:

lukranit [14]2 years ago

8 0

I think that it is 37

mylen [45]2 years ago

3 0

Answer:

7581, sorry if wrong

Step-by-step explanation:

9.5 times 9.5 = 90.25

90.25 times 12 = 1083

1083 times 7 = 7581

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?/0.6=1.4 how to show the solution

den301095 [7]

To find the solution to this problem, you would do the opposite of division which is multiplication.

Use the terms you have to plug into your new equation;

0.6 x 1.4 = .84

To check your work you would plug in .84 where the '?' is;

.84/.6= 1.4

There you have the original equation you began with.

Therefore, .84 would be your final answer.

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2 years ago

Giving Brainliest!! 2. Suppose that a frequency histogram and a cumulative frequency histogram are constructed from the same set

Ostrovityanka [42]

The answer would be:
B. The frequency is LESS THAN OR EQUAL TO the cumulative frequency.

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3 years ago

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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

$(A \cup B) + C = 1$