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3 years ago

6

If I get paid $8.00 per hour and I also get paid time-and-a-half for overtime hours, how much would I get paid for each overtime

hour?

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer:

Step-by-step explanation:

po

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A line passes through the points (–3, –4) and (6, 2)

erastova [34]

9514 1404 393

Answer:

(3, 0)

Step-by-step explanation:

The line through these points intersects the x-axis at x=3.

The x-intercept is (3, 0).

_____

The x-intercept can be computed from ...

x-intercept = x1 -y1(x2 -x1)/(y2 -y1)

= -3 -(-4)(6-(-3))/(2-(-4)) = -3 +4(9/6) = -3 +6

x-intercept = 3

__

Here, it is simple enough to plot the points and see where the line crosses the x-axis.

5 0

3 years ago

Math question please show work for brainliest :)

Aleks04 [339]

It’s 1 I think!

Hope I helped!!! :) :D

4 0

4 years ago

At the start of a year, company XYZ's stock is $40 per share. At the end, the company's stock is $60 per share. What was the sto

e-lub [12.9K]

We have been given that at the start of a year, company XYZ's stock is $40 per share. At the end, the company's stock is $60 per share. We are asked to find the percent increase.

We will use percent increase formula to solve our given problem.

$\text{Percent increase}=\frac{\text{Final}-\text{Initial}}{\text{Initial}}\times 100\%$

$\text{Percent increase}=\frac{60-40}{40}\times 100\%$

$\text{Percent increase}=\frac{20}{40}\times 100\%$

$\text{Percent increase}=0.5\times 100\%$

$\text{Percent increase}=50\%$

Therefore, the stock price's rate of return was 50%.

4 0

4 years ago

Expected Value (50 points)

garri49 [273]

Let $W$ be the random variable representing the winnings you get for playing the game. Then

$W=\begin{cases}10-1=9&\text{if the dice sum is odd}\\5-1=4&\text{if the dice sum is 4 or 8}\\50-1=49&\text{if the dice sum is 2 or 12}\\-1&\text{otherwise}\end{cases}$

First thing to do is determine the probability of each of the above events. You roll two dice, which offers 6 * 6 = 36 possible outcomes. You find the probability of the above events by dividing the number of ways those events can occur by 36.

The sum is odd if one die is even and the other is odd. This can happen 2 * 3 * 3 = 18 ways. (3 ways to roll even with the first die, 3 ways to roll odd for the die, then multiply by 2 to count odd/even rolls)
The sum is 4 if you roll (1, 3), (2, 2), or (3, 1), and the sum is 8 if you roll (2, 6), (3, 5), (4, 4), (5, 3), or (6, 2). 8 ways.
The sum is 2 if you roll (1, 1), and the sum is 12 if you roll (6, 6). 2 ways.
There are 36 total possible rolls, from which you subtract the 18 that yield a sum that is odd and the other 10 listed above, leaving 8 ways to win nothing.

So the probability mass function for this game is

$P(W=w)=\begin{cases}\frac12&\text{for }w=9\\\frac29&\text{for }w=4\text{ or }w=-1\\\frac1{18}&\text{for }w=49\\0&\text{otherwise}\end{cases}$

The expected value of playing the game is then

$E[W]=\displaystyle\sum_ww\,P(W=w)=\frac92+\frac89-\frac29+\frac{49}{18}=\frac{71}9$

or about $7.89.

7 0

3 years ago

In a box of chocolates, 1/5 of the chocolates contain nuts.

uysha [10]

Answer:

<em>1:4</em>

Step-by-step explanation:

<u>Ratios and Fractions</u>

It's given 1/5 of the chocolates in a box contain nuts. The rest do not contain nuts. The portion that does not contain nuts is:

$1 - \frac{1}{5}=\frac{4}{5}$

We need to calculate the ratio of the number of chocolates that contain nuts to the number of chocolates that do not contain nuts.

This can be done by dividing 1/5 by 4/5:

$\displaystyle \frac{\frac{1}{5}}{\frac{4}{5}}$

Operating:

$\displaystyle \frac{1}{5}\cdot\frac{5}{4}=\frac{1}{4}$

Expressing it as a ratio: 1:4

6 0

3 years ago