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3 years ago

Mr. Edwards has a total of 28 kids in his class. There are 6 more boys than there are girls. Write a system

Mathematics

1 answer:

ratelena [41]3 years ago

5 0

Answer:

17 boys and 11 girls

Step-by-step explanation:

Create a system of equations where b is the number of boys and g is the number of girls:

b + g = 28

b = g + 6

Solve by substitution, by substituting the second equation into the first one:

b + g = 28

(g + 6) + g = 28

2g + 6 = 28

2g = 22

g = 11

So, there are 11 girls.

Since there are 6 more boys than girls, there are 17 boys.

So, there are 17 boys and 11 girls in the class.

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If (11, 13) is an ordered pair of the function F(x), which of the following is an

FinnZ [79.3K]

The inverse is the exact opposite so -11, -13 and it if was negative points to plot it would just be possible.

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3 years ago

Reduce fraction 27/36 lowest term

natta225 [31]

The lowest term is 3
---
4

hope this helps

3 0

3 years ago

Read 2 more answers

A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the

iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

$Event\ 1 = \{1,2,3,4,5,6\}$

Next, we list out the sample space of F

$Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\$

$6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

In (1): the sample space of E is:

$E = \{1\}$

So:

$P(E) = \frac{n(E)}{n(Event\ 1)}$

$P(E) = \frac{1}{6}$

In (2): the sample space of F is:

$F = \{5,5,5,5\}$

So:

$P(F) = \frac{n(F)}{n(Event\ 2)}$

$P(F) =\frac{4}{36}$

$P(F) =\frac{1}{9}$

For E and F to be independent:

$P(E\ and\ F) = P(E) * P(F)$

Substitute values for P(E) and P(F)

This gives:

$P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}$

$P(E\ and\ F) = \frac{1}{54}$

However, the actual value of P(E and F) is 0.

This is so because $E = \{1\}$ and $F = \{5,5,5,5\}$ have 0 common elements:

So:

$P(E\ and\ F) = 0$

Compare $P(E\ and\ F) = \frac{1}{54}$ and $P(E\ and\ F) = 0$ .

These values are not equal.

Hence: the two events are not independent

6 0

2 years ago

How to simplify please help

Luda [366]

Okay, the image is a little tough to see, but I believe it looks something like this:

(5² -10 / 17 - 3 * 4) -1

The first step [remember PEMDAS] is parentheses

Next comes Exponents:

(25 - 10 / 17 - 3 * 4) -1

Then Multiplication:

(25 - 10 / 17 - 12) - 1

Then Division [which we can't do yet since it isn't fully simplified], so we skip to Addition... nothing again, then subtraction.

(15 / 5) - 1

Then we can divide:

(3) - 1

Then subtract!

The answer to your problem is 2.

Hope that helps!

4 0

3 years ago

What is the expanded form of 3x(x+3)^2?

Dominik [7]

We need to expand the square binomial and then multiply:
<span>3x(x + 3)^2
= 3x(x + 3)</span>(x + 3<span>)
= 3x(x^2 + 6x + 9)
= 3x^3 + 18x^2 + 27x</span>

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3 years ago