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2 years ago

At my favorite computer store, there was a sale of 20% off. I bought a $300

Mathematics

1 answer:

pochemuha2 years ago

5 0

<h2>Answer:</h2><h2>The total cost of the computer = $ 252</h2>

Step-by-step explanation:

The cost of the computer = $300

The sale percentage of the computer = 20%

20% ( 300) = $60

The current price of the computer = 300 - 60 = $240

5% of salex tax on the computer = 5% (240) = 12

The total price = current price of the computer + sales tax

= 240 + 12

= $252

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Quadrilateral A B D C is shown. In quadrilateral ABDC, AB ∥ CD. Which additional piece of information is needed to determine tha

Anna71 [15]

Step-by-step explanation:

For any quadrilateral to be a parallelogram

i) either both the pairs of opposite sides must be equal

ii) Both the pairs of opposite sides must be parallel

iii) Opposite pairs of angles must be equal

iv) Diagonals must bisect each other.

v) A pair of opposite sides must be parallel and equal

Here we are already given that AB || CD

So either we should be given that AD || BC

or we must be given that AB = CD

Here it is given AB = CD as an option.

So Option A) or the first option is the right answer that AB ≅CD is needed to prove ABCD is a parallelogram

3 0

2 years ago

Read 2 more answers

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

3 years ago

Solve for each System of Equation

weeeeeb [17]

using substitution.

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