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4 years ago

It takes 3.5 ounces of cocoa

Mathematics

1 answer:

schepotkina [342]4 years ago

5 0

Answer:

24.5 oz

Step-by-step explanation:

1 cup ----> required 3.5 oz of cocoa

1 cups -----> requires 3.5 oz x 7 = 24.5 oz of cocoa

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What is the answer I really need help

attashe74 [19]

Answer: 27

Step-by-step explanation:

The perimeter is the lengths of all sides of the figure added together. Since all sides are equal in this triangle, we know all sides have a length of 9. We know this because all sides have 2 little lines on them, symbolizing that they are equal.

P=9+9+9=27

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4 years ago

Find the surface area??

Dennis_Churaev [7]

The surface area is

1020.22

6 0

3 years ago

For the equation 3x - 5y = -3, what is the value ofy when x is 1? 6/5 -6/5 5/6

Law Incorporation [45]

Answer:

y = 6/5

Step-by-step explanation:

3x - 5y = -3

Let x =1

3(1) - 5y = -3

3-5y = -3

Subtract 3 from each side

3-3-5y = -3-3

-5y = -6

Divide each side by -5

-5y/-5 = -6/-5

y = 6/5

8 0

3 years ago

Read 2 more answers

Cedric opened a new gallon jug of milk and poured ¾ cup of milk for his mom, ½ cup milk for his little sister and 2 cups of milk

julia-pushkina [17]

Answer:

12 _3/4

Step-by-step explanation:

3/4 + 1/2 + 2 = 3_1/4

16 cups in a gallon

16-3_1/4

=12_3/4

7 0

4 years ago

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

3 years ago