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Alex Ar [27]
3 years ago
15

Which is not equivalent to the others please help

Mathematics
2 answers:
Sladkaya [172]3 years ago
4 0
The first one. So, 0.801
PtichkaEL [24]3 years ago
3 0

Answer:0.801

Step-by-step explanation:

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Find the vertex, focus, and directrix. y = 1/24(x+1)² - 3. 
Vedmedyk [2.9K]
y = \frac{1}{24}(x+1)^2 - 3\\\\y+3 =\frac{1}{24}(x+1)^2\ \ / *24\\\\ (x+1)^2 = 24(y+3)

This   is  an  equation  of  a  parabola  that  opens  upwards.

Its \ standard \ form: \\(x-h)^2=4p(y-k)\\ (h,k)=(x,y) \ coordinates \ of \ the \ vertex\\\ (h,k)=(-1,-3) \\\\axis \ of \ symmetry: \ x= -1\\ \\4p=24\ \ /:4\\p=6

focus:(h,k+p)=(-1,-3+6)=(-1,3) \\ \\directrix: \ y=k-p=-3-6=-9


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The movie rental store CineStar offers customers two choices. Customers
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Answer:

The yearly fee + $2 per movie

Step-by-step explanation:

Because lets say you rent 50 movies per year. Well with the yearly fee its $145. But with no yearly fee and just $3.50 per movie it’s $175.

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3 years ago
If pqr= 1 show that 1/(1+p+q^-1)+1/(1+q+r^-1)+1/(1+r+p^-1) = 1.<br>​
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Answer:

1

Step-by-step explanation:

1/(1+p+q^-1)+1/(1+q+r^-1)+1/(1+r+p^-1) = 1.

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Is 21/2+(-1/2)=-3 true?
natta225 [31]

Answer:

It's False!

Step-by-step explanation:

4 0
3 years ago
It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele
kow [346]

Answer:

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

\bar X=23500 represent the sample mean  

s=3900 represent the sample standard deviation      

n=100 sample size      

\mu_o =2000 represent the value that we want to test    

\alpha 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 determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:\mu \leq 2000      

Alternative hypothesis:\mu > 2000      

We don't know the population deviation, so for this case 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{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

df=n-1=100-1=99  

Since is a one-side upper test the p value would be:      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

8 0
3 years ago
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