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Verizon [17]
3 years ago
10

The measure of the vertex angle of an isosceles triangle is 15 more than the measure of each base angle. Find the degree measure

of each angle of the triangle

Mathematics
1 answer:
Strike441 [17]3 years ago
5 0
Answer:
Step by step explanation:

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A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has b
baherus [9]

Answer:

Null hypothesis:\mu \leq 6  

Alternative hypothesis:\mu > 6  

t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2    

df=n-1=36-1=35

t_{crit}=1.690 with the excel code:"=T.INV(0.95,35)"

t_{crit}=1.306 with the excel code:"=T.INV(0.90,35)"

p_v =P(t_{(35)}>2)=0.0267  

If we compare the p value and the significance level given \alpha=0.05,0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

Step-by-step explanation:

Data given and notation  

\bar X=6.5 represent the mean time for the sample  

s=1.5 represent the sample standard deviation for the sample  

n=36 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05,0.1 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 6 days, the system of hypothesis would be:  

Null hypothesis:\mu \leq 6  

Alternative hypothesis:\mu > 6  

If we analyze the size for the sample is > 30 but 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".  

Calculate the statistic

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

t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2    

Critical value and P-value

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

df=n-1=36-1=35

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates \alpha on the right. Using the significance level of 0.05 we got:

t_{crit}=1.690 with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

t_{crit}=1.306 with the excel code:"=T.INV(0.90,35)"

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

p_v =P(t_{(35)}>2)=0.0267  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05,0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

6 0
3 years ago
What is x-4y=28 2x+y=2 on a graph
trasher [3.6K]
<span>-4y + x = 28
<span>  y + 2x = 2

hope this helped :)
alisa202</span></span>
3 0
3 years ago
Someone please help me with this ASAP
Natasha_Volkova [10]

Answer:

y= -11

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
5(6.2+z) , z=3.8 help please :)
inna [77]

Answer:

the answer should be 50

Step-by-step explanation:


3 0
3 years ago
A sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's p
charle [14.2K]

Answer:

So the value of the test statistic is -0.85.

Step-by-step explanation:

We know that a sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 54% of the chips do not fail in the first 1000 hours of their use.

We get that:

n=1800\\\\p_1=53\%=0.53\\\\p_2=54\%=0.54\\

We calculate the standar deviation:

\sigma=\sqrt{\frac{0.53 \cdot (1-0.53)}{n}}=\sqrt{\frac{0.53 \cdot 0.47}{1800}}=0.01176

We calculate the value of the test statistic:

z=\frac{p_1-p_2}{\sigma}=\frac{0.53-0.54}{0.01176}=-0.85

So the value of the test statistic is -0.85.

5 0
3 years ago
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