Evaluate 0.1m+8-12n0.1m+8−12n0, point, 1, m, plus, 8, minus, 12, n when m=30m=30m, equals, 30 and n=\dfrac14n= 4 1 n, equals,
PIT_PIT [12445]
Answer:
8
Step-by-step explanation:
The task is to evaluate:
0.1m + 8 - 12n
When 
By substituting these values into the expression, we have:

Step One
Deduct 32 from both sides.
F - 32 = \frac{9}{5}(k - 273.15)
Step Two
Multiply each side by \frac{5}{9}.
\frac{5}{9}(F - 32) = \frac{5}{9} \times \frac{9}{5}(k - 273.15)
\frac{5}{9}(F - 32) = k - 273.15
Step Three
Add 273.15 to both sides.
\frac{5}{9}(F - 32) + 273.15 = k
Problem B
F = 180
Solve for k
k = \frac{5}{9}(F - 32) + 273.15
k = \frac{5}{9}(180 - 32) + 273.15
k = \frac{5}{9} \times 148 + 273.15
k = 82.2222 + 273.15
k = 355.3722
k = 355.4 <<< Answer
<span>Determine the configuration of columns and rows for the rectangular arrangement of 120 cupcakes.
=> There must be an even number of rows and an odd number of columns.
=> 120 = 2 x 2 x 2 x 15
=> 120 = 8 x 15
=> 120 = 120
Consequently, the glee club should organize the cupcakes in 8 rows and 15 columns.
This totals up to 120 cupcakes altogether.
</span>
Answer:
To inspect a batch consisting of 20 semiconductor chips, a sample of 3 is selected. Out of these, 10 chips fail to meet customer specifications.
a) Total distinct samples possible = 20C3 =
=1140
b) For exactly 2 good chips and 1 bad chip
Total samples = 10C2 * 10C1 = 45 * 10 =450
c) Combinations of 2 good 1 bad, 1 good 2 bad, and 3 bad chips
Total samples = 10C2 * 10C1 + 10C1 * 10C2 + 10C3
= 
I think the right answer is D. 8x+8;
This is because Frank possesses X stamps while Chris has 8 plus 7 times the number of stamps Frank holds, making the total 8 + 7x (which is 7 times Frank's amount), so when you add Frank's stamps (x) to 8 + 7x, it results in 8 + 8x.
I hope this was helpful!