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nalin [4]
2 years ago
12

Susan is going to watch a movie in her collection she has 5 action movies 4comedies and 3 dramas she will randomly select one mo

vie what is the probability tagt the movie she selects is not drama hurry please
Mathematics
1 answer:
Ivenika [448]2 years ago
6 0

Answer:

9/12 or 75%.

Step-by-step explanation:

Let's first add the choices all up.

5+4+3= 12.

We have 12 movie choices in total.

Now, we know that we have 3 drama movie choices...so the chance we will choose drama is 3/12, but we are not finding that.  We want the chance she will NOT choose drama, so that would be 9/12.

*Since the chance she will choose drama is 3/12, the chance she will not choose is 9/12. 9+3 is 12, the total amout.*

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The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
timurjin [86]

Answer:

a) Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

b) p_v =2*P(t_{(24)}  

c) If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The values obtained are:

\bar X=85272 represent the mean annual salary for the sample  

s=11039.23 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =90000 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:  

Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(24)}  

Part c: Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

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3 years ago
Is 0.9 greater,least than, or equal to 95%​
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Answer:

Less

Step-by-step explanation:

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Answer:

B) AD/DE = BD/AD

Step-by-step explanation:

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5. Two unbiased dice are rolled. Calculate the probability that the sum of the two dice is:
IceJOKER [234]
The sample space has 36 possible pairs from 1,1 1,2 1,3 up to 6,5 and 6,6

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(c) 15 pairs add to less than 7 so P(<7) = 15/36 = 5/12
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3 years ago
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soldier1979 [14.2K]
So.. 2.86 - 2.79 is the amount in decrease or 0.07

so.. if we take 2.86 as the 100%, how much is 0.07 of that in percentage?

well \bf \begin{array}{ccllll}&#10;amount&\%\\&#10;\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\&#10;2.86&100\\&#10;0.07&x&#10;\end{array}\implies \cfrac{2.86}{0.07}=\cfrac{100}{x}

solve for "x"
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