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defon
3 years ago
12

Probability ----------------

Mathematics
1 answer:
NISA [10]3 years ago
6 0

Step-by-step explanation:

P(W^c) = 1-3/5

=5-3/5 =2/5

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Determine what the solution is to the following system? Y=3/4x-1 & y=4/3x-1
vichka [17]

Answer:

(0, -1)

Step-by-step explanation:

There are multiple ways of solving this however- since both equations are already in Y-Intercept form, we will use the "Equal Values Method"

First, since both equations are equal to Y, we can set them equal to each other and solve for X

\frac{3 }{4} x - 1 =  \frac{4}{3} x - 1

To start, you must eliminate the fraction using a "fraction buster" multiply EVERYTHING by 4 then simplify.

3x - 4 = 5 \frac{1}{3}x  - 4

Since we still have a fraction, we shall do it one more time. This time we multiply by 3

9x - 12 = 6 x - 12

Now, solve how you normally would.

9x = 6x

-6x

3x = 0

X = 0

Now, since we know what X would equal in the solution, we are able to plug in X as 0 in one of our equations. We can choose the first one!

y =  \frac{3}{4} (0) - 1

Now solve which would lead to y = -1

You have your solution as

(X,Y)

(0,-1)

Hope this helps!

4 0
3 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

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 is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 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 true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

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: P-value  and conclusion

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

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

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 is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
What is the measure of NMR? <br><br> 27<br> 63<br> 107<br> 117
ASHA 777 [7]
63 because both angles are the same
5 0
3 years ago
If the graph of the following parabola is shifted two units left and three units down, what is the resulting equation?
mafiozo [28]

Answer:

x=-8\left(y+3\right)^2-2

Step-by-step explanation:

Because the x and y have been reversed in this equation, shifting the parabola to the left will be outside of the exponent and shifting up and down will be inside with the exponent.

The equation becomes x=-8\left(y+3\right)^2-2.

Where subtracting by 2 moves it two units to the left and adding by 3 moves it 3 units down.

7 0
3 years ago
Hurry, it’s a test!!! Plz!
kondor19780726 [428]

Answer:

estimate 215 answer 10

Step-by-step explanation:

6 0
3 years ago
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