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3 years ago

If a cookie weighs

Mathematics

2 answers:

slamgirl [31]3 years ago

8 0

If you do all the math which is pretty simple you would have a total weight of 288 ounces or 18 pounds.

m_a_m_a [10]3 years ago

5 0

1 cookie = 3/4 of an ounce

16 cookies = ?

3/4 x 16 = 12

16 cookies = 12 ounces

24 packages = ?

12 x 24 = 288

24 packages = 288 ounces of cookies.

So, your final answer is 288 ounces.

I hope this helps! Good luck!

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An isosceles triangle has two identical angles.

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3 years ago

Read 2 more answers

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

lubasha [3.4K]

Answer:

a. 10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. 0.2457 = 24.57% probability 10 of the problems can be resolved today

c. 0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today

d. 0.5017 = 50.17% probability more than 10 of the problems can be resolved today

Step-by-step explanation:

For each case, there are only two possible outcomes. Either they are solved, or they are not. Cases are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

It can resolve customer problems the same day they are reported in 80% of the cases.

This means that $p = 0.8$

Sample of 13 cases:

This means that $n = 13$

a. How many of the problems would you expect to be resolved today? What is the standard deviation?

Expected:

$E(X) = np = 13*0.8 = 10.4$

Standard deviation:

$\sqrt{V(X)} = \sqrt{13*0.8*0.2} = 1.44$

10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. What is the probability 10 of the problems can be resolved today?

This is $P(X = 10)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

0.2457 = 24.57% probability 10 of the problems can be resolved today.

c. What is the probability 10 or 11 of the problems can be resolved today?

This is $p = P(X = 10) + P(X = 11)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$p = P(X = 10) + P(X = 11) = 0.2457 + 0.2680 = 0.5137$

0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today.

d. What is the probability more than 10 of the problems can be resolved today?

This is

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$P(X = 12) = C_{13,12}.(0.8)^{12}.(0.2)^{1} = 0.1787$

$P(X = 13) = C_{13,13}.(0.8)^{13}.(0.2)^{0} = 0.0550$

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) = 0.2680 + 0.1787 + 0.0550 = 0.5017$

0.5017 = 50.17% probability more than 10 of the problems can be resolved today

6 0

3 years ago

On a circle, which of these is not a straight line?

Y_Kistochka [10]

Answer:

circumference

Step-by-step explanation:

it's a circle

6 0

3 years ago