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djyliett [7]
2 years ago
8

Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible

by 5. Note: The digit sum of 1440 is 1+4+4+0 = 9.​
Mathematics
1 answer:
alisha [4.7K]2 years ago
6 0

Answer:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999

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Step-by-step explanation:


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A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?
nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

                P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean SAT math score for in-state applicants = 540

\bar X_2 = sample mean SAT math score for out-of-state applicants = 555

s_1 = sample standard deviation for in-state applicants = 20

s_2 = sample standard deviation for out-of-state applicants = 25

n_1 = sample of in-state applicants = 35

n_2 = sample of out-of-state applicants = 35

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }  = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means (\mu_1-\mu_2) is ;

P(-1.997 < t_6_8 < 1.997) = 0.95  {As the critical value of t at 68 degree

                                         of freedom are -1.997 & 1.997 with P = 2.5%}  

P(-1.997 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.997) = 0.95

P( -1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

P( (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

<u>95% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

=[(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } },(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

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Answer:

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Step-by-step explanation:

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If you have variables on both sides of the equation, pick one of the variable terms and subtract it from both sides of the equation.

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If we choose to subtract x, then we will have a variable term on the left and a  constant term on the right:

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  x = 1 . . . . . . simplify

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Note that we purposely set up this example so that removing the variable term from the right side caused the variable term and constant term to be on opposite sides of the equal sign. It may not always be that way. As long as you remember that an unwanted term can be removed by subtracting it (from both sides of the equation), you can deal with constant terms and variable terms no matter where they appear.

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<em>Additional Comment</em>

It usually works well to choose the variable term with the smallest (or most negative) coefficient. That way, when you subtract it, you will be left with a variable term that has a positive coefficient.

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