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sladkih [1.3K]
3 years ago
11

Two decimal numbers have a difference of 0.7 And and a sum of 1.What are the numbers?

Mathematics
2 answers:
Andreas93 [3]3 years ago
4 0
Use an equation for questions like these:
I'll use x and y as the unknown numbers
x-y=0.7
x+y=1
Just choose either x or y to solve for
y=x-0.7
y=-x+1
Then since since they're both equal to y then they can equal each other
x-0.7=-x+1
2x-0.7=1
2x=1.7
x=0.85
Then just grab one of the equations and use it to find y
0.85+y=1
y=0.15

So the two numbers are 0.85 and 0.15
klio [65]3 years ago
3 0
The two numbers are 0.85 and 0.15
if you add them together you get 1, 0.85+0.15=1
and if you subtract them you get 0.7, 0.85-0.15=0.7
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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.  

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