Answer:
z
Step-by-step explanation:
the same time as the first time since
A two-digit number is twice the sum of its digit. If the tens digit is 7 less than the unit digit, find the number.
Let x= the unit digit
Then y= the tens digit
<span>And 10y+x= the number
</span>x+y= the sum of the digits
<span>Now we are told that 10y+x=2(x+y) ------1st equation </span>
<span>We are also told that y=x-7 ----------- 2nd Equation </span>
<span>So our equations to solve are: </span>
(1) 10y+x=2(x+y)
<span>(2) y=x-7
</span>
Hope it helps
I believe you get 10 dollars in interest.
Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get 
we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.
The outlier (61) is at the low end of the data set, but doesn't affect the mean by a lot, so ...
The mean is centered among the other numbers in both sets of data.
_____
The mean without the outlier is 114. With the outlier, it is 107.4. The lower quartile is 108, so the mean does get moved outside the "box" of the box-and-whisker plot of the data set without the outlier.
