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3 years ago

You bought a large cooler that holds 5 gallons. How many quarts is that?

2 answers:

6 0

Your 5 gallon cooler would equal to 20 quarts
There are 4 quarts in a gallon and there are 5 gallons here so it would be 4 x 5 which equals 20 quarts would fit in the cooler

Korvikt [17]3 years ago

4 0

First we will see how many quarts are in a gallon.

1 gallon = 4 quarts.

Now lets MULTIPLY to get our answer.

5 gallon = ? quarts

5 × 4 = 20

So, the cooler has 20 quarts or 5 gallons.

Hope I helped ya!!!!!!!

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Select the correct answer.

EleoNora [17]

Answer:

Option (B)

Step-by-step explanation:

There are two lines on the graph representing the system of equations.

First line passes through two points (-3, 1) and (-2, 3).

Slope of the line = $\frac{y_2-y_1}{x_2-x_1}$

= $\frac{3-1}{-2+3}$

m = 2

Equation of the line passing through (x', y') and slope = m is,

y - y' = m(x - x')

Equation of the line passing through (-3, 1) and slope = 2 will be,

y - 1 = 2(x + 3)

y = 2x + 7 ----------(1)

Second line passes through (0, 1) and (-1, 4) and y-intercept 'b' of the line is 1.

Let the equation of this line is,

y = mx + b

Slope 'm' = $\frac{y_2-y_1}{x_2-x_1}$

= $\frac{4-1}{-1-0}$

= -3

Here 'b' = 1

Therefore, equation of the line will be,

y = -3x + 1 ---------(2)

From equation (1) and (2),

2x + 7 = -3x + 1

5x = -6

x = $-\frac{6}{5}$

x = $-1\frac{1}{5}$

From equation (1),

y = 2x + 7

y = $-\frac{12}{5}+7$

= $\frac{-12+35}{5}$

= $\frac{23}{5}$

= $4\frac{3}{5}$

Therefore, exact solution of the system of equations is $(-1\frac{1}{5},4\frac{3}{5})$ .

Option (B) will be the answer.

5 0

3 years ago

Iq scores are normally distributed with a mean of 100 and a standard deviation of 15.

bearhunter [10]

Mean is $\mu=100$ and a standard deviation is $\sigma =15$ , then the variable $X\sim N(100,15^2)$ .

Use substitution

$Z=\dfrac{X-\mu}{\sigma},\\ \\ Z=\dfrac{X-100}{15}$ .

This substitution gives you that $Z\sim N(0,1)$ .

a. For X=130, $Z=\dfrac{130-100}{15}=2$ and $Pr(X>130)=Pr(Z>2) =0.9772$ (the decimal value is taken from the Standard Normal Distribution Table).

b. For X=90, $Z=\dfrac{90-100}{15}=-\dfrac{2}{3}$ and for X=110, $Z=\dfrac{110-100}{15}=\dfrac{2}{3}$ . Then $Pr(90$ (the decimal value is taken from the Standard Normal Distribution Table).