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Elena L [17]
3 years ago
6

If a and b are both positive, two-digit integers, is a + b a multiple of 11? The tens digit of a is equal to the units digit of

b, and the tens digit of b is equal to the units digit of a Both a and b are odd.
Mathematics
1 answer:
Serhud [2]3 years ago
6 0

Answer:

Yes!

Step-by-step explanation:

Let x be the tens place and y be the units place. x and y need to be odd because in other case a or b will not be odd. For example, if x=1 and y=2 a will be 21 but b will be 12 that is not odd.

Now, a+b = xy+yx. Note that xy is not x*y, is just the digits concatenated.Then, there are two cases:

If x+y<10 then xy+yx = (x+y)(x+y) (again, that is not a multiplication is x+y concatenated with x+y) and that is 11*(x+y) a multiple of 11.

If x+y≥10 then xy+yx = (x+y+1)(x+y-10) because x+y<20.

Now we are going to see that last result without the concatenation, as a sum, that is

(x+y+1)*10 + x+y-10 = 10x+10y+10+x+y-10 = 11x+11y = 11(x+y). This result is clearly a multiple of 11.

In conclusion, in all cases the result of a+b is a multiple of 11.

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Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

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

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 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 check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

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

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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