Helpppppp fastttttttttt

of the 75 boys in the 7th grade class, 25 participate in at least one sport . of the 120 girls in 7th grade class 30

hoa [83]

Answer:

"No, the relationship is NOT proportional"

Step-by-step explanation:

<u>Complete Question:</u>

Of the 75 boys in the 7th grade class, 25 participate in at least one sport. Of the 120 girls in 7th grade class 30 participate in at least one sport. Is this relationship proportional?

We have to see the ratio of each and if they are equal (after reducing), we can say they are proportional.

Boys:

25 participate in atleast 1 sport and there are total of 75. Hence, ratio is

25 : 75 = 1 : 3

Girls:

30 particiapte in atleast 1 sport and there are toal of 120. hence ratio is:

30 : 120 = 1 : 4

So, the relationship is NOT proportional

6 0

3 years ago

The sales tax rate is 8.2%. Estimate the sales tax on a computer that costs $550.

Butoxors [25]

Asking the Math Gods...

550*.082+550=$595.10

$45.10 in tax

7 0

3 years ago

Read 2 more answers

Help plz show work I need to have all Aś this year Only do 18,20,24,22,28,26,30,32,

Veseljchak [2.6K]

Answer: see below

<u>Step-by-step explanation:</u>

The "point" on the inequality symbol points to the smaller number

< is the "less than" inequality symbol
> is the "greater than" inequality symbol
= is the "equal" symbol

18) 3 = 3

20) $\sqrt6 < \sqrt{10}$

22) 0.8 = 4/5 Note: 4÷5 is 0.8

26) 0.075 < 0.39

28) -5.2 < -4.8

30) 0.4 > $\sqrt{0.4}$

32) $\sqrt5 < \sqrt7$

6 0

2 years ago

Please Help Me it's due soon

anyanavicka [17]

-1 and 4 I think is the answer is the function

4 0

2 years ago

A= [2, 3; 2, 1], B=[3; 5]. Find AB & BA if possible

oksano4ka [1.4K]

Answer:

The value of AB is $\left[\begin{array}{ccc}21\\11\end{array}\right]$ and it's not possible to multiply BA.

Step-by-step explanation:

Consider the provided matrices.

$A=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right]$ , $B=\left[\begin{array}{ccc}3\\5\end{array}\right]$

Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.

Multiply AB

Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:

$AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]$

$AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]$

$AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]$

$AB=\left[\begin{array}{ccc}21\\11\end{array}\right]$

Hence, the value of AB is $\left[\begin{array}{ccc}21\\11\end{array}\right]$

Now calculate the value of BA as shown:

Multiply BA

Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.

We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.

That means number of column of first matrix should be equal to the number of rows of second matrix.

Hence, it's not possible to multiply BA.

5 0

3 years ago