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storchak [24]
3 years ago
9

You are making a colorful cake for a birthday party and need to use 4 drops of food coloring for every 5 ounces of icing. How ma

ny drops of food coloring do you need if
you are using 30 ounces oficing?
Mathematics
1 answer:
kari74 [83]3 years ago
4 0

Answer:

24 drops of food coloring

Step-by-step explanation:

First, since there are 30 ounces, divide by 5, because there are 4 drops in every 5 ounces of icing. Since 30÷5=6, You have to multiply it by 4 to get 24. You multiply by 4 because there are 4 drops in 5 ounces of icing.

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A man that is 6 feet tall is standing so that the tip of his shadow is 20 feet from a light pole. His shadow is 8 feet long. Wha
pychu [463]

Answer:

The answer to your question is: the height of the light pole is 15 ft.

Step-by-step explanation:

See the picture below

Now, we can do proportions to finds the height of the light pole

                   \frac{20}{8} =  \frac{x}{6}

Solve for x

                      x = 6(20) / 8

                     x = 120 / 8

                     x = 15 ft

                   

4 0
3 years ago
Researchers fed mice a specific amount of Toxaphene, a poisonous pesticide, and studied their nervous systems to find out why To
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Answer:

a. The mean refractory period= 1.85 and the standard error = 0.06455

b.  90% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment = 1.6981, 2.0019

c. Yes, the data give good evidence to support this theory

Step-by-step explanation:

a.  The table below shows the calculations:

                 X                (X-mean)^2

                1.7             0.0225

                1.8                 0.0025

                1.9                 0.0025

                2.0                 0.0225

Total        7.4                  0.05

Sample size: n=4

The mean is:  \bar{x} = \frac{7.4}{4} = 1.85

The sample standard deviation, s = \sqrt{\frac{\sum \left ( x-\bar{x} \right )^{2}}{n-1}}=0.1291

The standard error, se= \frac{s}{\sqrt{n}}=\frac{0.1291}{2} = 0.06455

b. Degree of freedom: df = n-1 = 3

Critical value of t for 90% confidence interval is: 2.3534

The confidence interval is  \bar{x}\pm t_{c}se = 1.85\pm 2.3534\cdot 0.06455=1.85\pm 0.1519 = (1.6981, 2.0019)

c. The Hypotheses are:

H_{0}:\mu=1.3,H_{1}:\mu>1.3

So the test statistics will be

t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=8.52

The p-value is: 0.0017

We reject the null hypothesis because p-value is less than 0.05 . This indicates that the data gave good evidence to support this theory.

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Thats a reasonable price for a sweater.
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Is there a none option? if not then it has to be B.
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