The sum of the digits of a certain two-digit number is 7. Reversing its digits increase the number by 9. What is the number?

2 answers:

5 0

Let the two numbers be x and y.

According to your question;

x + y = 7
10y + x = 10x + y + 9

By equation 1 ; x = 7-y

Substituting the value of x ;

10y + ( 7 -y) = 10(7-y) + y + 9

9y + 7 = 70 -10y + y + 9

9y + 7 = 70 - 9y + 9

=> 18y = 70 -7 + 9
=> 18y = 72
=> y = 4

Substituting for x ;

x = 7 - y
=> x = 7 -4
=> x = 3

Thus, x = 3 and y = 4;
=> The number is 34.

AleksandrR [38]3 years ago

3 0

The original number is <u>34 </u>.
The sum of its digits is (3 + 4) = 7 .
Reversing its digits makes it 43 . . . 9 more than 34.

How I did it: (There must be a much more elegant mathematical method)

There are only 6 possibilities for the original number.
They are . . .

16
25
34
43
52
61 .

Instead of trying to dream up a sophisticated equation,
I decided it was faster and easier to just go down the list
and look for one that worked.
The third one I tried worked.

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.

umka21 [38]

Answer:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

$\bar X_1 = 28900$ the sample mean for the salaries of teachers in Indiana

$s_1 = 2300$ the sample deviation for the salary of teachers in Indiana

$n_1 =10$ the sample size from Indiana

$\bar X_2 = 30300$ the sample mean for the salaries of teachers in Michigan

$s_2 = 2100$ the sample deviation for the salary of teachers in Michigan

$n_2 =14$ the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom for this case are:

$df = n_1 +n_2 -2= 10+14-2 =22$

The confidence is 95%so then the significance is $\alpha=0.05$ and the $\alpha/2 =0.025$ , we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got: