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Deffense [45]
3 years ago
13

Mrs. Reid is going on a trip. She has 9 book that she hasn't ready yet, but she wants to bring only 3 on the trip.

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0
The way you work this out is by thinking about the odds of each singular event, then finding the overall odds based on the individual odds.

The number of different books Mrs. Reid can choose is 9, so the first number is 9.

Mrs. Reid has picked one book so far, so now she has (1 - 9) = 8 books to choose from.
The of different books Mrs. Reid can choose now is 8, so your second number is 8.

Mrs. Reid has picked 2 book thus far, so now there are (2 - 9) = 7 books to choose from.
The of different books Mrs. Reid can choose now is 7, so your second number is 7.

To get the total number of different choices, multiply all the singular events together:

7*8*9=504\ different\ ways\ to\ bring\ 3\ books
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3 years ago
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Vanyuwa [196]

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8 0
2 years ago
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Kyle received the following course grades for 10​ (three-credit) courses during his freshman​ year:
timofeeve [1]

Answer:

(a) The average grade point is 2.5.

(b) The relative frequency table is show below.

(c) The mean of the relative frequency distribution is 0.3333.

Step-by-step explanation:

The given data set is

4, 4, 4, 3, 3, 3, 1, 1, 1, 1

(a)

The average grade point is

Average=\frac{\sum x}{n}

Average=\frac{4+4+4+3+3+3+1+1+1+1}{10}

Average=2.5

Therefore the average grade point is 2.5.

(b)

\text{Relative frequency}=\frac{\text{Frequency of number}}{\text{Total frequency}}

The relative frequency table is show below:

x                 f                  Relative frequency

4                3                 \frac{3}{10}=0.3

3                3                 \frac{3}{10}=0.3

1                 4                 \frac{4}{10}=0.4

(c)

Mean of the relative frequency distribution is

Average=\frac{0.3+0.3+0.4}{3}

Average=0.3333

Therefore the mean of the relative frequency distribution is 0.3333.

4 0
2 years ago
Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv
Georgia [21]

Answer:

Since our calculated value is lower than our critical value,z_{calc}=2.02, we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation  

\bar X_{1}=5.2 represent the mean for group men  

\bar X_{2}=4.5 represent the mean for group women  

Assuming these values for the remaining data:

\sigma_{1}=1.2 represent the population standard deviation for the sample men

\sigma_{2}=1.5 represent the population standard deviation for the sample women

n_{1}=32 sample size for the group men  

n_{2}=30 sample size for the group women  

z would represent the statistic (variable of interest)

p_v represent the p value  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:  

H0:\mu_{1} \leq \mu_{2}  

H1:\mu_{1} > \mu_{2}  

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

3) Calculate the statistic  

We have all in order to replace in formula (1) like this:  

z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02  

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got z_{critical}=2.33

5) Statistical decision

Since our calculated value is lower than our critical value,z_{calc}=2.02, we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0
3 years ago
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