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Korvikt [17]
4 years ago
11

Diana made a recipe that yields 6 and 1 over 2 cups. If each serving is 1 over 4 cup, which expression will help Diana determine

the number of servings her recipe will yield? 6 and 1 over 2 ⋅ 1 over 4 6 and 1 over 2 ÷ 1 over 4 6 and 1 over 2 + 1 over 4 6 and 1 over 2 − 1 over 4
Mathematics
2 answers:
Misha Larkins [42]4 years ago
6 0

Answer:

6 and 1 over 2 ÷ 1 over 4

Step-by-step explanation:

Given,

The total cups in a recipe =  6 and 1 over 2 cups

=6\frac{1}{2}\text{ cups}

Also, the cups in each serving =  1 over 4 cup

=\frac{1}{4}\text{ cups}

Hence, the number of servings = \frac{\text{total cups in a recipe}}{\text{cups in each serving}}

=\frac{6\frac{1}{2}}{\frac{1}{4}}

= 6 and 1 over 2 ÷ 1 over 4

Second option is correct.

Degger [83]4 years ago
5 0
Six and a half divided by one over four
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I cant see the image...but I take it that the midpoint is (9,8) and the endpoint S is (10,10) and ur looking for the other endpoint R.

midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
(10,10)....x1 = 10 and y1 = 10
(x,y)....x2 = x and y2 = y
so we sub
(10 + x) / 2, (10 + y) / 2 = 9/8

(10 + x) / 2 = 9
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The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10
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For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

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So the value of mode is also 4.5338 G.

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So Variance = (0.1039)^2

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To learn more about normally distribution link is here

brainly.com/question/15103234

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