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3 years ago

How to do the fractions

Mathematics

1 answer:

GaryK [48]3 years ago

3 0

First you make the mixed numbers/fractions into improper fractions. After that, you must find the common denominator, the LCM of 9 and 6 (for the 2nd problem). After that, if you can, simplify the fraction to lowest terms, which also includes making it a mixed number/fraction.

I hope this helps! :)

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To graph the line y = 3/2x + 7 *

Firdavs [7]

Answer:

Answer is C

Step-by-step explanation:

Hope it helps!

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3 years ago

Jasmine's dog weighs 15kg the dogs collar weighs 200 g. how many grams does the dog weigh when it wearing the collar?

Cerrena [4.2K]

There are 1000 grams in a kg - so the dog weighs 15,000 grams. Add that to the collar weight (200g) and you get 15,200 grams.

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3 years ago

What value is to 18 - 3(4-6)*2

Ilya [14]

Answer:

Step-by-step explanation:

18-3(4-6)*2

=18-3*(-2)*2

=18-3*(-4)

=18-(-12)

=18+12

=30

8 0

4 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

The ratio of 6th graders to 7th graders in a MATHCOUNTS club is 5:4. The ratio of 8th graders to 7th graders in the same club is

lorasvet [3.4K]

ANSWER
========

Ratio of 8th graders to 7th graders = 3 : 2

Let the number of graders be 3x and 2x

Number of 8th graders = 18

So , 3x = 18

x = 18/3 = 6

Number of 7th graders = 2 × 6 = 12

Ratio of 6th graders to 7th graders = 5 : 4

Let the number of graders be 5a and 4a

Number of 7th graders = 12

So , 4a = 12

a = 12/4 = 3

Number of 6th graders = 5×3 = 15 ANS

_______________________________

#BE BRAINLY

3 0

3 years ago

Read 2 more answers