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2 years ago

You buy a game for $ 40 , and the sales tax is $2 . What percent of the game price are you paying in sales tax?

2 answers:

5 0

So you are buying a video game that is worth $40 and you had to use the sales tax, with is an extra $2, so the percentage would be 5% since 5% = $2 and when you finish filling the percentage to 100%, it would be 100% = $40.

Hope this helped!

Nate

Rina8888 [55]2 years ago

3 0

Answer: 5% or .05

Step-by-step explanation:

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Evaluate 5c+cd when c=1/5 and d=15

oee [108]

5*1/5+1/5*15=4
1 + 3 =4

This should equal 4 if I did this correctly.

4 0

3 years ago

Lcm of (x-2)(x+3) and 10(x+3)^2

Arte-miy333 [17]

Least common multiple: factor them, then see what they have in common and what is leftover and multiply those expressions:

(x - 2)(x + 3) 10(x + 3)(x + 3)
Common: (x + 3)
Leftover: (x - 2), (10), (x + 3)
Common · Leftover is: (x + 3) · (x - 2) · (10) · (x + 3) = 10(x - 2)(x + 3)²

Answer: LCM is 10(x - 2)(x + 3)²

8 0

2 years ago

The price of a particular stock is represented by the linear equation y=-0.91x+103.47

Mama L [17]

To find the price of the stock after 12 weeks, you would have to plug in 12 to x

That would make your equation: y = -0.91(12) + 103.47

Then, you solve:

-0.91(12) would be -10.12

Your equation will now look like this:

y = - 10.12 + 103.47

Then you add -10.12 to 103.47

-10.12 + 103.47 = 92.55

y = 92.55

After 12 weeks, the price of the stocks would be $92.55

4 0

3 years ago

Read 2 more answers

Someone help with this

elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

2 years ago