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Mashcka [7]
4 years ago
6

At Stephanie's new job, she spent $6.50, $8, $7.25, $13.50, and $9.75 on lunch the first week. In the second week, Stephanie spe

nt a total of $10 less than the total of the first week. What is the decrease in the mean for the second week compared to the first week?
answer as a decimal rounded to the nearest hundredth.
Mathematics
1 answer:
MaRussiya [10]4 years ago
6 0

Answer:

The mean for the second week is $2 less than the first and in percentage it is 22% less.

Step-by-step explanation:

The mean is given by the sum of all individual values divided by the number of values. For the first week the sum is:

sum1 = 6.5 + 8 + 7.25 + 13.5 + 9.75

sum1 = 45

Since she spent 10 less in the second week the sum is:

sum2 = sum1 - 10 = 45 - 10 = 35

The mean for each week is:

mean1 = sum1/5 = 45/5 = 9

mean2 = sum2/5 = 35/5 = 7

difference = mean1 - mean2 = 9-7 = 2

difference(%) = [2/9]*100 = 0.22*100 = 22%

The mean for the second week is $2 less than the first and in percentage it is 22% less.

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Need help on probability. Please explain how you did it so I can know how to do it.
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Answer:

\text{P}(B \text{ or }C)=\dfrac{5}{6}

Step-by-step explanation:

<u>Mutually Exclusive Events</u>

For two events, A and B, where A and B are mutually exclusive:

\boxed{\text{P}(A \text{ or }B)=\text{P}(A)+\text{P}(B)}

<u>Probability distribution table</u>:

\begin{array}{|c|c|c|c|c|c|c|c|}\cline{1-7} x & 1 & 2 & 3 & 4 & 5 & 6 \\\cline{1-7} \text{P}(X=x)\phantom{\dfrac11}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\\cline{1-7}\end{array}

where X is the score on a fair, six-sided dice.

P(B or C) means "the probability of rolling an even number or a multiple of 3".

As an even number of a fair, six-sided dice can never be a multiple of 3, the two events B and C are mutually exclusive.

Calculate the probabilities for events B and C.

<u>Event B</u>

Rolling an even number.

\begin{aligned}\implies \text{P}(X \text{ is even})&=\text{P}(2 \text{ or }4\text{ or }6)\\\\&=\text{P}(X=2)+\text{P}(X=4)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{3}{6}\end{aligned}

<u>Event C</u>

Rolling a multiple of 3.

\begin{aligned}\implies \text{P}(X \text{ is multiple of 3})&=\text{P}(3 \text{ or }6)\\\\&=\text{P}(X=3)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{2}{6}\end{aligned}

<u>Solution</u>

\begin{aligned}\implies \text{P}(B \text{ or }C)&=\text{P}(B)+\text{P}(C)\\\\& = \dfrac{3}{6}+\dfrac{2}{6}\\\\& = \dfrac{5}{6}\end{aligned}

6 0
2 years ago
Read 2 more answers
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