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Vlada [557]
3 years ago
10

Marcus baked a loaf of banana bread for a party. He cut the loaf into equal size pieces. At the end of the party, there 6 pieces

left. Explain how you can find the number of pieces in the whole loaf if Marcus told you that 1/3 of the loaf was left.
Mathematics
1 answer:
hram777 [196]3 years ago
4 0
Marcus is left with 6 pieces and 1/3 of the loaf. so  by the problem if x is the whole loaf 1x/3 =6 x =18 . The total pieces of loaves  were 18
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Preston is 8 years less than twice as many years old as tyler. If tyler is 12 years old, how old is preston
tia_tia [17]

Answer:

P=2T-8, if T=12, Preston, P=24-8=16 years old.

Hope this helps ;)

7 0
2 years ago
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Find the value of x in each case:
diamong [38]

Answer:

Step-by-step explanation:

Sum of 3 angles of triangle= 180

147 + 4x + x = 180

147 +5x = 180

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3 years ago
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Circle the volume in cm^3 of a cylinder with the radius 5cm and height 8cm
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Answer:

<em>628 </em>cm^{3}

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Since the formula to find the volume of a cylinder is \pi r^{2}h, we just need to fill things in.

Since the radius is 5 and the height is 8 and we know that pi=3.14, our new equation is:

3.14*5^2*8.

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3.14 * 25 * 8

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Hope this helps!! <3 :)

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3 years ago
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If (x + 1)(x - 3) = 12, then which of the following statements is true?
Alexxandr [17]
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4 0
3 years ago
Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank
mars1129 [50]

Complete question:

Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.

Answer:

11/288

Step-by-step explanation:

We are given:

Box 1: ( 4White, ORed)

Box 2: (3White, 1Red)

Box 3: (2White, 2 Red)

We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.

Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.

Our Event, E = getting 2 red balls.

Now Box 1 is ruled out, we have:

P[E(B1)]= 0

P[E/B3)] = (2 2) / (4 2)

= 1/6

If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:

P(E/B2) = (1/4) *(1/4)

= 1/16

Therefore probability that 2 balls extracted are red, we have:

P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 0 * \frac{1}{3} + \frac{1}{6}*\frac{1}{3} + \frac{1}{16}*\frac{1}{3}

= 11/288

3 0
3 years ago
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