All categories

3 years ago

Which of the following could be the numbers of left-handed and right-handed students at West Junior

Mathematics

1 answer:

nordsb [41]3 years ago

3 0

Answer:

The anwers Is C and E.

Step-by-step explanation:

A relationship between two quantities is proportional if the ratio between those quantities is always equivalent.

Hint #22 / 4

Let's look at the ratio between right- and left-handed students at East Middle School.

right

left

right per left

58left

609right

=10.5 right per left

At East Middle School, there are

10.5 right-handed students for every left-handed student.

Hint #33 / 4

Now we can compare the other combinations of values for left-handed and right-handed students. Which ratios are equivalent to the ratio at East Middle School?

So the anwers is C AND E

You might be interested in

What goes into the square root of 96?

VladimirAG [237]

Hi there, the factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96. So, we want the highest factor of 96. The factor we will use is the number 16. Lets get solving. √96=√16*6=√16*√6=4√6, Therefore, the square root of 96 simplified is 4√6 and the square root of 96 is 9.79795897113

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!! PLEASE ANSWER A AND B! GIVING AWAY 13 POINTS

telo118 [61]

According to google A. Would be 2.375 or as a fraction it would be 2 3/8

5 0

3 years ago

Please help!! What Find the Error and explain why it is wrong!

Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. The real numerator in Step E should be:

$3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x$

Hence, Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

And further steps are described below:

Step D

$x^{2}-x+4 = 0$