All categories

4 years ago

Noel has 5/6 of a yard of purple ribbon and 9/10 of a yard of pink ribbon. How much ribbon does she have altogether?

Mathematics

2 answers:

Rom4ik [11]4 years ago

4 0

You get 14/16 and when you simply you then get 7/8.

Andrews [41]4 years ago

4 0

To do this, you will want to make both of the denominators common or in other words, the same. The easiest way is to multiply the denominators together.
(5*10)/(6*10) and (9*6)/(10*6)
50/60 and 54/60
Add these together and you get 104/60 which is equal to 1.7333... yards of ribbon.

Hope I helped!

You might be interested in

Solve - 9/2y=27/2 for y.

dedylja [7]

Answer:

It’s y = 3

Step-by-step explanation:

Solve for y by simplifying both sides of the equation, then isolating the variable.

5 0

3 years ago

How to do this question plz answer me step by

Jet001 [13]

Answer:

481.92

Step-by-step explanation:

First find the increase

466.98 * 3.2%

466.98 * .032

14.94336

Add this to the original amount

466.98+14.94336

481.92336

Round to 2 decimal places

481.92

7 0

3 years ago

In a Create of 30 eggs<br>only 60 % is good how many<br>are bad?<br><br>

andrew11 [14]

Answer:

In my point of view 40% are bad

5 0

3 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

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