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Svetlanka [38]
3 years ago
12

The total amount of Keira's cell phone bill, b, varies directly with the number of text messages she sends, m. Which of the foll

owing equations best models the amount of money Keira must pay for her cell phone bill?
Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0
Pls list the equation options so we can give you an answer
You might be interested in
The longer base of a trapezoid is 8 ft. The longer base of a similar trapezoid is 13 ft. The area of the smaller trapezoid is 24
Licemer1 [7]

Answer:

390 ft²

Step-by-step explanation:

The longer base of a trapezoid is 8 ft. The longer base of a similar trapezoid is 13 ft. The area of the smaller trapezoid is 240 ft² What is the area of the larger trapezoid?

We solve the above question using proportion

(Longer base/Area of trapezoid) smaller trapezoid = (Longer base/Area of trapezoid) bigger trapezoid

Let the the Area of the bigger trapezoid = x

Hence,

= 8ft/240ft = 13ft/x ft

Cross Multiply

8ft × x = 240ft × 13ft

x = 240ft² × 12 ft/8 ft

x = 390 ft²

3 0
2 years ago
The length of a rectangle is 3 more than 3 times its width. The perimeter of the rectangle is 174 inches. What is the length of
saveliy_v [14]

Answer:

<em>l = w + 3cm</em>

<em>l = w + 3cmp = 2l + 2w = 58cm</em>

<em>l = w + 3cmp = 2l + 2w = 58cm </em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 </em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 Plug back in:</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 Plug back in:l = (13cm) + 3cm = 16cm</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 Plug back in:l = (13cm) + 3cm = 16cmStep-by-step explanation:</em>

I hope this helps you.

5 0
2 years ago
Evaluate each expression for the given value (please answer all)
Anna [14]

Answer:

-20a - 65 = -104  for  a = 2.7

-6x + 52 = 45  for  x = 1 1/6

Step-by-step explanation:

-20a - 50  for  a = 2.7

Insert 2.7 in for a.

-20(2.7) - 50 = -54 - 50 = -104

-6x + 52  for  x = 1 1/6

Insert 1 1/6 for x.

-6(1 1/6) + 52 = -7 + 52 = 45

8 0
3 years ago
What is the least common multiple of the numbers 5,25, and 15
liubo4ka [24]

Answer:

5

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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
2 years ago
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