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Alja [10]
3 years ago
15

if my goal is 27,729,350 and i get 150 closer every minute and 15 seconds how long is it going to take me to get to my goal

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer: 13,864,675 seconds

Step-by-step explanation:

I minute 15 seconds = 75 seconds

That means it takes 75 seconds to achieve 150 , and it takes x seconds to achieve 27,729,350. Interpreting this in proportion form , we have:

150 ------ 75 seconds

27,729,350 ------- x seconds

Then

150x = 27,729,350 X 75

150x = 2,079,701,250

x =  2,079,701,250 / 150

x = 13,864,675

That means it takes 13,864,675 seconds to achieve the goal , converting this to minutes, we have:

231,077.9 minutes

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Answer:

Opposite coordinate of (−5, −7) = (1,1)

Step-by-step explanation:

Let the sides of rectangle be a and b

Perimeter = 2 ( a+b ) = 28

                       a + b = 14

Area = ab = 48

The sides with sum 14 and product 48 is 8 and 6.

a = 8 and b = 6

Since origin is inside the circle, the side with 8 unit is parallel to y axis and side with 6 unit is parallel to x axis

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An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha
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Answer:

t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

Step-by-step explanation:

Data given and notation  

\bar X=6.3 represent the sample mean    

s=1.0 represent the sample standard deviation

n=26 sample size  

\mu_o =6.0 represent the value that we want to test  

\alpha=0.025 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 ture mena is higher than 6.0, the system of hypothesis would be:  

Null hypothesis:\mu \leq 6.0  

Alternative hypothesis:\mu > 6.0  

Since we don't know the population deviation, 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{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

P-value  

The degrees of freedom are given by:

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

Conclusion  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

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