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olga2289 [7]
3 years ago
12

Hey does somebody think they can help me find the value of this expression?

Mathematics
2 answers:
tatiyna3 years ago
4 0
The answer would be -2/1
——
4
MissTica3 years ago
3 0

Answer:

-\frac{3}{4}

Step-by-step explanation:

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Help me please. I’ll mark brainliest
Dvinal [7]

Answer:

I don't really understand this question

But if it's asking you to switch,

1. r = 2.5

2. r = 12

3. d = 34

6 0
3 years ago
a suit cost $95 Barry has a discount coupon for 10% off and the city has a 4.5 percent sales tax what is the final cost of the s
SIZIF [17.4K]
SO the suit cost $95 and it has a discount of 10%
so 95*10/100=9.5
so 10% of 95 is 9.5
subtract 9.5 from 95 and that equals 85.5
then you have to add 4.5 percent sales tax
so 85.5+ 4.5%= 89.347
So the final cost would be $89.34
3 0
3 years ago
Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa
blagie [28]

Answer:

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

p_v =P(z>1.9)=0.0287  

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 12 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=12.95 represent the sample mean

\sigma=3 represent the population standard deviation for the sample  

n=36 sample size  

\mu_o =12 represent the value that we want to test  

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

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:\mu \leq 12  

Alternative hypothesis:\mu > 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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".  

Calculate the statistic  

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

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>1.9)=0.0287  

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 12 at 5% of signficance.  

4 0
3 years ago
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Answer:

See I don't know the Answer but I Need points to ask question so Sorry

Step-by-step explanation:

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