Simplify each equation,then solve

Mathematics

1 answer:

Colt1911 [192]3 years ago

3 0

I would need the equations

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Hailey spent $12 to buy 2 fish. Her cousin spent double the amount on fish. Each fish cost the same amount. How many fish do Hai

Greeley [361]

Haley spent $ 12....her cousin spent double....so her cuz spent $ 24

haley spent $ 12 for 2 fish...12/2 = 6 bucks per fish
so if the cuz spent $ 24....24/6 = 4.....cuz got 4 fish

haley's 2 fish + cuz's 4 fish = 6 fish in total

3 0

3 years ago

Read 2 more answers

The daily earnings of an employee who works on a commission basis are given by the following probability distribution. Find the

elena-s [515]

Answer: The employee's expected earnings is 78.5.

Step-by-step explanation:

Since we have given that

$ Probability

0 0.07

25 0.12

50 0.17

75 0.14

100 0.28

125 0.18

150 0.04

We need to find the employee's expected earnings.

As we know the formula for "Expectation":

$E(x)=0\times 0.07+25\times 0.12+50\times 0.17+75\times 0.14+100\times 0.28+125\times 0.18+150\times 0.04\\\\E(x)=0+3+8.5+10.5+28+22.5+6\\\\E(x)=78.5$

Hence, the employee's expected earnings is 78.5.

7 0

3 years ago

A slice is made perpendicular to the base of a right rectangular prism as shown.

Gala2k [10]

The cross section is just a rectangle. So to work out the area, multiply the length of the base together by the height.
This is 22 x 12 = 264cm²

3 0

3 years ago

Suppose you bought 20 tickets in a drawing for a color television. There were 300 tickets sold in the drawing,

grandymaker [24]

1/15 is your probability

7 0

2 years ago

determine the quadratic function of f whose graph is given. The vertex is (1,-3) and the y-intercept is -2

LuckyWell [14K]

Hmmm the y-intercept is at -2? what does that mean? well, is where the graph "intercepts" or touches the y-axis, and when that happens, x = 0, so the point is really ( 0 , -2 ).

and we know where the vertex is at. Let's assume a vertical parabola, in which case the squared variable is the "x".

$\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
vertex
\begin{cases}
h=1\\
k=-3
\end{cases}\implies y=a(x-1)^2-3
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=0\\
y=-2
\end{cases}\implies -2=a(0-1)^2-3
\\\\\\
1=a(-1)^2\implies \boxed{1=a}\qquad thus\implies 
\begin{cases}
y=1(x-1)^2-3\\
\textit{or just}\\
y=(x-1)^2-3
\end{cases}$

6 0

3 years ago