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3 years ago

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A pancake recipe calls for 1/3 cup of mix for 4 pancakes. If Beth needed to make 60 pancakes, how many cups of pancake mix will

she need?
A:5 cups
B:4 2/3 cups
C:3 1/3
D:1/5

2 answers:

zheka24 [161]3 years ago

7 0

The answer is A, Im pretty sure it is.

just olya [345]3 years ago

5 0

1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3

4 4 4 4 4 4 4 4 4 4

I think its A

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What’s the equation for this table? Use y=mx+b. HELP

Verdich [7]

Answer:

y = 2x + 12

Step-by-step explanation:

You can check this by plugging any of the values from the table into the equation. You can find the answer by finding the slope and then using that to find the y-intercept.

y2-y1/x2-x1

16-14/2-1 = 2/1 = 2. So our slope equals 2.

If we set x = 0 to find our y-intercept, and we know that we have a slope of 2 we can subtract 2 from 14 to get the y-intercept.

14-2 = 12. --> (0, 12) which is true because we know that the slope is 2.

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3 years ago

A holiday decoration involves blue and white lights the ratio of blue lights to white lights is three to two the box the decorat

Westkost [7]

Answer:

15

Step-by-step explanation:

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3 years ago

What is the area of a cross section that is parallel to face CDHG ?

Dennis_Churaev [7]

Answer: Area of cross section that is parallel to face CDHG is 432 cm².

Step-by-step explanation:

Since we have given that

There is a cross section that is parallel to face CDHG.

So, Length of cross section would be 36 cm

Width of cross section would be 12 cm.

So, Area of cross section would be

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8 0

3 years ago

Read 2 more answers

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago

A glacier that was 1,076 meters thick change in thickness at an average rate of -22.7 meter per year for 7 years.

-Dominant- [34]

1,076 (the original thickness) - 22.7 meters (loss per year) * 7 (lost 22.7 meters per year for seven years, so multiply the loss per year by the amount of years) = 917.1 (the size after 7 years)

A.K.A 1,076 - 22.7 * 7 = 917.1

i found this off of the internet so im sorry if its wrong

6 0

3 years ago