Its asking how much did they both spent, in total, in order to reach the same amount spent in part A
Well, following the order of PEMDAS, I got choice B. 52
For instance, when you plug in 5 for x, you get F(5)=2(5)^2+2.
Moreover, following PEMDAS, you're supposed to solve what's inside the parenthesis, but since there is no operation going on inside the parenthesis, then you simple move on to the exponent.
In this case, you square the number 5, which gives you F(5)=2(25)+2
After that, you Multiply (letter M in PEMDAS). This results in F(5)=50+2.
Finally, you add them, which results in F=52.
By the way, I noticed a mistake in your work. When multiplying 2 by 5, the answer is 10, not 20.
Anyway, hope this helped! :-)
Answer:
Option C: n = 32; p^ = 0.4
Step-by-step explanation:
The normal curve can be used in this case if; np ≥ 10 or n(1 - p) ≥ 10
A) For n = 28 and p = 0.3;
np = 28 × 0.3 = 8.4 < 10
Thus, it can't be used.
B) For n = 28 and p = 0.9;
np = 28 × 0.9 = 25.2 > 10 Ok
n(1 - p) = 28(1 - 0.9) = 2.8 Not Ok
Thus, it can't be used
C) For n = 32 and p = 0.4
np = 32 × 0.4 = 12.8 > 10 Ok
n(1 - p) = 32(1 - 0.4) = 19.2 > 10 Ok
Thus, it can be used
D) For n = 32 and p = 0.2
np = 32 × 0.2 = 6.4 < 10 Not Ok
Thus it can't be used.
Answer:
AB = 4
BC = 4
AC = 8
Step-by-step explanation:
Given, B is the midpoint of AC which means it divided the AC into two equal parts. Using this information we can write the following equation:
5x - 6 = 2x add 6 to both sides
5x = 2x + 6 subtract 2x from both sides
3x = 6 divide both sides by 3
x = 2 to find the length of AB, AC, and BC replace x with 2
AB = 5*2 - 6
BC = 2*2
AC = 8 (sum of AB and BC)
Answer: The quotient = 10
Step-by-step explanation: A quotient is the result you get when a number is used to divide another number. e . g.
6¹/₄ ÷ ⁵/₈ ( where 5/8 is the divisor )
First convert 6¹/₄ to improper fraction
= 25/4 ÷ 5/8
= 25/4 × 5/8
= 10
The quotient = 10.
NOTE: The 5/8 changed to 8/5 because when dividing fractions, immediately you applied the multiplication symbol, the denominator of the divisor becomes the numerator and vice versa
