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timurjin [86]
3 years ago
12

You need 3 sticks of butter for every 24 cookies you bake. How many sticks of butter for 5 cookies

Mathematics
2 answers:
tigry1 [53]3 years ago
5 0

Answer:

5:40

Step-by-step explanation:

24 / 3 = 8

x / 5 = 8

x = 5 * 8

x = 40

xxTIMURxx [149]3 years ago
4 0

Answer:

you will need 5/8 sticks  of butter for 5 cookies

Step-by-step explanation:

<u><em>You need 3 sticks of butter for every 24 cookies you bake. How many sticks of butter for 5 cookies</em></u>

<u><em /></u>

To solve this, you simply need to use proportion;

3 sticks = 24 cookies

 x          = 5 cookies

cross multiply

24 × x  = 3 × 5

24x = 15

Divide both-side of the equation  by 24 in order to get the value of x.

24x/24  = 15/24

( on the left-hand side of the equation, 24  at the numerator will cancel-out the 24 at the denominator  and we will be left with just x, while on the right- hand-side of the equation, 15 will be divided by 24)

x = 15/24

(3 can divide both 15 and 24, so we will reduce 15 and 24 to its lowest term by dividing the two numbers by three).

x =  5/8

Therefore, 5/8 stick of butter is needed for 5 cookies

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k=57

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Multiply both sides by 5

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-10 + -5 HEEEEEEEEEELP
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Find the 11th term of the arithmetic sequence 10,-40,160,….
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3 years ago
Lizzie rolls two dice. What is the probability that the sum of the dice is:
zhenek [66]

Answer:

A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A)  = \dfrac{12}{36} = \dfrac{1}{3}

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)}  Count 1

Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

P(B)  = \dfrac{15}{36} = \dfrac{5}{12}

C. Divisible by 3 and less than 7:

P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

P(A \cap B) = \dfrac{7}{\text{36}}

5 0
3 years ago
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