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Lelu [443]
4 years ago
6

Easy points!!! Correct answer gets brainliest. :)

Mathematics
2 answers:
zzz [600]4 years ago
8 0
Hello!

The correct answer is C. 2.5 To figure this out, we can multiply each answer choice by x, or (4, 8, 12).

Hope this helps! ☺♥


Anna007 [38]4 years ago
7 0
I'll vouch for quiverer and tell you to give him the brainliest!!!
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What are the factors of -12?
Alenkinab [10]

Answer:

− 12  ,  − 6 , − 4 , − 3, −2, −1, 1, 2, 3, 4, 6 , 12

Step-by-step explanation:

-12 * 1 = -12

-6 * 2 = -12

-4 * 3 = -12

-3 * 4 = -12

-2 * 6 = -12

-1 * 12 = -12

1 * -12 = -12

2 * -6 = -12

3 * -4 = -12

4 * -3 = -12

6 * -2 = -12

12 * -1 = -12

8 0
3 years ago
Does 4y = 2y-12 have a solution? Please include step by step to help me understand and solve future problems
Ray Of Light [21]

Answer:

Hope its helped you

Step-by-step explanation:

4 0
3 years ago
You are to play the following game of cards. Cards are worth their face value, Jacks, Queens and Kings are also worth 10 and Ace
ryzh [129]

Answer:

$82.875

Step-by-step explanation:

From the given information:

Assume F is used to denote the two cards;

If there are four aces among 52 playing cards, the chance of selecting the first ace is =  \dfrac{4}{52}

After selecting the first ace, we have only 3 aces remaining and a total of 51 playing cards. Thus, the chance of selecting the second ace will be \dfrac{3}{51}

By applying product rule, we can determine the chance of selecting two aces without replacement as follows:

i.e.

P(F=22)=\dfrac{4}{52}\times \dfrac{3}{51}

= \dfrac{1}{221}

The probability of getting one ace, one face is:

P(F=21) =( \dfrac{4}{52}\times \dfrac{12}{51})+( \dfrac{12}{52}\times \dfrac{4}{51})

P(F=21) = \dfrac{4}{221}\times\dfrac{4}{221}

P(F=21) = \dfrac{8}{221}

Since there are 4 aces, 4 nine, and 12 faces in a card deck

The probability of getting one ace, one nine, or two faces now will be:

P(F=20) = (\dfrac{4}{52} \times \dfrac{4}{51})+ (\dfrac{4}{52} \times \dfrac{4}{51}) + (\dfrac{12}{52} \times \dfrac{11}{51})

P(F=20) = (\dfrac{53}{663}  )

Now, the probability  of at least 20 now is:

\text{P(F at least 20)} = \dfrac{1}{221}+\dfrac{8}{221}+\dfrac{53}{663}

\text{P(F at least 20)} = \dfrac{80}{663}

If H represents the amount of prize of the expected winnings:

Then;

(H - 10) (\dfrac{80}{663}) + (-10)(\dfrac{663-80}{663}) = 0

\dfrac{80(H-10)}{663}-\dfrac{5830}{663}=0

\dfrac{80(H-10)}{663}=\dfrac{5830}{663}

80H - 800 = 5830

80H = 5830 +800

80H = 6630

H = 6630/80

H = $82.875

The prize should be $82.875 to make a winning positive.

4 0
3 years ago
10. What is the value of x in the figure at the right? (Examples 3 and 4
Montano1993 [528]

Answer:

x = 37

Step-by-step explanation:

Vertically opposite angles are equal

Therefore,  2x + 6 = 80

Subtract 6 from both sides:   2x = 74

Divide both sides by 2:   x = 37

8 0
3 years ago
Read 2 more answers
What number much each side of the equation 2/5 x=10 be multiplied by to produce the equivalent equation x=25
iris [78.8K]
2/3x=10
3/2 x 2/3x = 3/2 x 10
x=30/2=15
7 0
3 years ago
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