9.25 the average score for Mrs Myles class was 70% of Mrs Jones i class test average score in Mr. Myles class

Fofino [41]

To first find the answer to this, let's find a common denominator between each fraction. The denominator is the bottom number of a fraction.
To find the common denominator, let's factor out each of our denominators 5 times to see if they have a common number.
4: 4, 8, 12, 16, 20
2: 2, 4, 6, 8, 10
They both appear to have the smallest common denominator of 4, so let's use 4.

We already have 1/4 with a denominator of 4, but not 1/2.
2 x 2 = 4, so multiply the numerator and the denominator of 1/2 by 2.
1 x 2 = 2
2 x 2 = 4
Your new fraction is 2/4.

Our expression now is:
-2 1/4 + 5 2/4
or (in simpler terms)
5 2/4 - 2 1/4

Subtract the fractions first.
2/4 - 1/4 = 1/4

Now subtract the whole numbers.
5 - 2 = 3

You are left with:
3 1/4

I hope this helps!

8 0

2 years ago

Point P is shown on the number line below. The distance between point Q and point P is 6 1/2 units. Which number could represent

Bad White [126]

B is the right answer

6 0

2 years ago

Read 2 more answers

Pls I need help on this problem

Liono4ka [1.6K]

Answer:

x = 64

Step-by-step explanation:

they give you two angles 56 and 60

add them and subtract from 180

180 - 116 = 64

7 0

3 years ago

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

jasenka [17]

Answer:

The minimum score of those who received C's is 67.39.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 73, \sigma = 11$