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creativ13 [48]
3 years ago
5

What is $70 discounted to $63 in percent form

Mathematics
1 answer:
RideAnS [48]3 years ago
3 0
First step:

difference: $70 - $63 = $7

Second step:
\frac{\$7}{\$70}=\frac{1}{10}=0.1

Third step:
0.1\times100\%=10\%

Answer: Discounted is equal 10%.
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8 times the sum of a number and 9 is -32
Andre45 [30]

Answer:

Answer:

1)

2)

3) The greatest age that Sue could be is 7.

4) The smaller of the two integers is 92.

5) Don needs to earn at least 244 points in the fourth game.

Step-by-step explanation:

1) An rectangle has 2 dimensions: width(w) and length(l)

The perimeter P is:

The problem states that the length of a rectangle is three times its width. So l = 3w and:

The perimeter of the rectangle is at most 112 cm. It means that the perimeter can be 112, so the equal sign enters the inequality. So

2)

The problem states that he earns $850 per week in sales. His earnings is modeled by the following equation:

, in which w is the number of weeks.

The problem also states that he spent $7500 to obtain his merchandise, and it costs him $300 per week for general expenses. So his expenses can be modeled by the following equation

, in which w is also the number of weeks.

He will make a profit when his earnings are bigger than his expenses, so: When they are equal, there is no profit, so the equal sign does not enter the inequality.

3)

I am going to call Jenny's age x and Sue's age y.

The problem states that Jenny is eight years older than twice her cousin Sue’s age. So

.

The sum of their ages is less than 32, so:

Sue's age has to be less than 8, so the greatest age that Sue could be is 7.

4)

The sum of two consecutive integers is at least 185.

There are two integers with sum of 185, so:

They are consecutive so:

Replacing in the sum equation:

The smaller of the two integers is 92.

5)

The average is the sum of all the scores divided by the number of games. So:

Don needs to earn at least 244 points in the fourth game.

6 0
3 years ago
The sum of two numbers is 48. If one third of one number is 5 greater than one sixth of another number, which of the following i
lubasha [3.4K]
<span>The sum of two numbers is 48.
a + b = 48
;
If one third of one number is 5 greater than one sixth of another number,
a = b + 5
multiply both sides by 6, cancel the fractions
2a = b + 30
2a - b = 30
</span><span>use elimination to solve this
a + b = 48
2a - b =30
-------------Addition eliminates b, find a
3a = 78
a = 
a = 26
then
26 + b = 48
b = 48 - 26

b = 22</span>
5 0
3 years ago
Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of
steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

  • From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack)  P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
  • A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

  • WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the  probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

  • WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a  king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been  removed.

  • WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick  a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

  • The probability of drawing the second queen is also  4/52 and the third is 4/52.
  • We multiply these three individual probabilities together to get P(QQQ) =
  • P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
  • Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
5 0
3 years ago
Which of the following equations shows how substitution can be used to solve
maxonik [38]

Answer:  Choice B)

2x+5(2x+3) = 3

============================================

Reasoning:

We start with the second equation 2x+5y = 3

Since y = 2x+3 from the first equation, we can replace every copy of y with (2x+3). This replacement is valid because y and 2x+3 are the same.

So that's how we go from 2x+5y = 3 to 2x+5(2x+3) = 3

This replacement is done to get rid of the y variable, and it allows us to solve for x later on.

3 0
3 years ago
Find the two numbers whose sum is 54 and whose difference is 4
galina1969 [7]

Answer:

25 and 29

Step-by-step explanation:

25 + 29 = 54 and there is a difference of 4 between them

4 0
3 years ago
Read 2 more answers
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