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alekssr [168]
3 years ago
12

Pamela and her friend Nicole are each baking apple pies and tarts for a bake sale, using the same recipes. Pamela baked 5 apple

pies and 7 apple tarts, using a total of 68 apples. Nicole made 5 apple pies and 9 apple tarts, which used 76 apples. How many apples does each dessert require?
Mathematics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

Each apple pie requires 8 apples, and each apple tart requires 4 apples.

Step-by-step explanation:

We see that both Pamela and Nicole bake the same amount of apple pies, but different amounts of apple tarts. Because of this, we can subtract the two to try to figure out the amount of apples for each apple tart. We subtract 68 from 76, giving us 8. Nicole baked 9 apple tarts, while Pamela baked 7, and 9-7=2. So we can bake two apple tarts with 8 apples, so one apple tart requires 4 apples (we divide by 2). Now that we know the amount of apples per each apple tart, we multiply 7 apple tarts that Pamela made by 4 apples, giving us 28. We subtract that from the total amount of apples Pamela used, which was 68, giving us 40. From this we can deduct that 5 apple pies need 40 apples, and we divide by 5, giving us 1 apple pie requires 8 apples.

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B8%7D" id="TexFormula1" title="\frac{5}{8}" alt="\frac{5}{8}" align="absmiddle
bija089 [108]

Answer:

\frac{5}{12}

Step-by-step explanation:

Multiply straight across and reduce.

\frac{5}{8} · \frac{2}{3} = \frac{10}{24} = \frac{5}{12}

6 0
3 years ago
A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible o
Agata [3.3K]

Answer:

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

Step-by-step explanation:

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG.

Let X represent the number of children that are girls. Then

1. When X=0, there is one possible outcome BBBB. So

Pr(X=0)=\dfrac{1}{16}

2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so

Pr(X=1)=\dfrac{4}{16}=\dfrac{1}{4}

3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so

Pr(X=2)=\dfrac{6}{16}=\dfrac{3}{8}

4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so

Pr(X=3)=\dfrac{4}{16}=\dfrac{1}{4}

5. When X=4, then there is one possible outcome GGGG, so

Pr(X=4)=\dfrac{1}{16}

Now, the probability distribution table is

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

5 0
4 years ago
Read 2 more answers
Name tree pairs of numbers that have 5 as the greatest common factor (GCF) and use each number once
podryga [215]
(15, 5)
(10, 30)
(20, 25)
6 0
4 years ago
Erin had 55 stuffed bears. She took out her favorite 7 bears and then equally divided the other bears among her 3 sisters. Erin'
Gemiola [76]
>.> love bears
55 - 7 = 48
48/3 = 16
16 bears each,
Su had 15 + 16
Su has 31 bears
Hope this helps
8 0
4 years ago
12,20,28,36 are all multiples of 1, 2 and what?
liberstina [14]

1) one is multiple of all positive integers. So, yes it is multiple of all that numbers.

2) Two is multiple for all those number which has even digit in unit place. So, yes they all are multiple of 2 too.

4 0
4 years ago
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