Answer:
C.
Step-by-step explanation:
sorry if im wrong
Hello there!
So, let's start off with the first one:
A) 60% of One - half x
I believe a couple of possible equations would be: 0.60*1/2x or 0.60*0.5x. 1/2 and 0.5 both equal half of a number. The 0.60 would equal 60%. I'll plug 10 into x of both equations to see what I come up with:
First Equation:
1/2 * 10/1 = 10/2 or 5.
0.60 * 5 = 3
Second Equation:
0.5 * 10 = 5
0.60 * 5 = 3
So they both come up with the same answer, so that makes them both equal, which makes them both possible equations.
B) 7 times z reduced by a third of the product
7z - 1/3 seems like a good equation. Let me plug in 5 for z to solve:
7 * 5 = 35
Convert 35/1 into 105/3 by multiplying both sides by 3
105/3 - 1/3 = 104/3 (approx. 34.7)
C) Three-fourths of V subtracted from 6 times two-ninths
A good equation for this would probably be (6*2/9)-3/4v. I'll plug 5 into v to show how we would solve the equation:
6/1 * 2/9 = 12/9 or 1 1/3.
3/4 * 5/1 = 15/4 or 3 3/4.
(Use the fractions 16/12 and 45/12 as 16/12 = 12/9 and 15/4 = 45/12)
16/12 - 45/12 = -29/12 or −<span>2 <span>5/12</span></span>
D) Sum of 2x, (2/3 + 5) and 11-x
I think the right equation for this would be 2x + (2/3 + 5) + 11-x. Plug in 3 for x to solve:
(Use 15/3 when adding 2/3 and 5 because 5 = 15/3)
2/3 + 15/3 = 17/3 or 5 2/3
2 * 3 = 6
11 - 3 = 8
(Use 18/3 and 24/3 when adding up 6 + 5 2/3 + 8 as 6 = 18/3 and 8 = 24/3
17/3 + 18/3 + 24/3 = 59/3 or 19 2/3
Sorry if this confused you xD but, basically, the possible equations are as follows:
1 = 0.60*1/2x or 0.60*0.5x
2 = 7z - 1/3
3 = <span>(6*2/9)-3/4v
</span>4 = 2x + (2/3 + 5) + 11-x
I hope this helped! If not, I'm very sorry. Algebra isn't my strong suit ^.^'
Answer:
18, 23
Step-by-step explanation:
a. We know that W = 5 and D = 3 so:
P = 3W + D = 3 * 5 + 3 = 18 points
b. We know that P = 50 and D = 14 and that we are solving for W so we get:
50 = 3W + 14
36 = 3W
W = 12 which means the losses are 35 - 12 = 23.
I believe it is this first one but if you get it wrong I am sorry
Fun. I prefer Oxymetazoline.
For the control group we have a headache probability of
c = 368/1671 = .220
For the experimental group we have a headache probability of
e = 494/2013 = .245
The observed difference is
d = e - c = .025
The variance of the difference is
s² = c(1-c)/n₁ + e(1-e)/n₂
so the standard deviation is
We get a t statistic on the difference of
t = d/s = .025/.0139 = 1.79
We're interested in the one sided test, P(d > 0). We have enough dfs to assume normality. We look up in the standard normal table
P(z < 1.79) = .96327
so
p = P(z > 1.79) = 1 - .96327 = 0.037 = 3.7%
Answer: That's less that 10% so we have evidence to conclude that headaches are significantly greater in the experimental group.
