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ASHA 777 [7]
1 year ago
15

Help for a brianlest at show your work

Mathematics
1 answer:
Nastasia [14]1 year ago
5 0
X^2-2(x-y)-z^3
First we substitute the variables for their known values and get
2^2-2(2-(-2))-(-3)^3
Simplify
4 + (-4+4) -(-27)
4+-27= -23
That’s your answer
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On a loan of $4,500 for 2 1/2 years at 4% per year, how much do you wind up paying to pay off the loan?
Anna35 [415]
4/100= x/4,500 18,000/100= $180x2.5= $450
7 0
3 years ago
Read 2 more answers
A car rental agency advertised renting a car for $24.95 per day and $0.26 per mile. If Brad rents this car for 3 days, how many
bulgar [2K]

Answer:

481 whole miles

Step-by-step explanation:

Total cost would be 3 x 24.95 + 0.26m, where m is the miles driven.

Total cost has to be under or equal to $200, so

3 x 24.95 + 0.26m <= 200

74.85 + 0.26m <= 200     ( 3 x 24.95)

0.26m <= 125.15               ( subtract 74.85)

m <= 481.35                      ( divide 0.26)

3 0
4 years ago
Read 2 more answers
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
I have 9000 grams of chocolate. I know that 1 pound is equal to 0.45 Kg. How many pounds of chocolate do I have?
Sphinxa [80]

Answer:

9

Step-by-step explanation:

7 0
3 years ago
Any body can help me like pls
Valentin [98]

Answer:

Option D

<h2>Answer is D- 1/5</h2><h2></h2><h2>Please mark as brainliest for future answers :)</h2>

6 0
3 years ago
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