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3 years ago

Chrissy ate 10% of the Christmas cookies. If she ate 6 cookies, how many Christmas cookies were left?

Mathematics

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

If Chrissy eats 6 cookies so 54 cokies are left. .

Step-by-step explanation:

You might be interested in

Is (5,3) a solution to this system of equation <br> X+y=8<br> 17x+17y=8

Degger [83]

(5, 3)

x y

Plug in:

x + y = 8

5 + 3 = 8

8 = 8

Plug in:

17(5) + 17(3) = 8

85 + 51 = 8

136 = 8

Answer - No, (5, 3) is not a solution to this system of equations.

5 0

3 years ago

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

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

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

1 year ago

A researcher was interested in comparing the salaries of female and male employees of a particular company. independent random s

Aneli [31]

Answer:

Based on given data, we can say with 95% confidence that the female employees at this company average between $110 less and $10 less per week than the male employees.

Step-by-step explanation:

Sample 1 for female

Sample 2 for male

Confidence = 90%

Difference in mean weekly salary of all female employees and the mean weekly salary of all male employees = µ₁- µ₂

So based on above given data, we can say with 95% confidence that the female employees at this company average between $110 less and $10 less per week than the male employees.

5 0

3 years ago

Calculate 6.75 tax on 23.05

Ksenya-84 [330]

The answer is 12dollas

3 0

2 years ago

Evaluate the problem <br>

m_a_m_a [10]

Answer:

-2

Step-by-step explanation: