Answer:
Step-by-step explanation:
Confidence interval for the difference in the two proportions is written as
Difference in sample proportions ± margin of error
Sample proportion, p= x/n
Where x = number of success
n = number of samples
For the men,
x = 318
n1 = 520
p1 = 318/520 = 0.61
For the women
x = 379
n2 = 460
p2 = 379/460 = 0.82
Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.025 = 0.975
The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96
Margin of error = 1.96 × √[0.61(1 - 0.61)/520 + 0.82(1 - 0.82)/460]
= 1.96 × √0.0004575 + 0.00032086957)
= 0.055
Confidence interval = 0.61 - 0.82 ± 0.055
= - 0.21 ± 0.055
Answer:answer from math papa is b=−n+30
pictures is blurry please send another
Your basically building a new equation with the two functions given to you.
(sqrt(3x + 7)) + (sqrt(3x - 7)) = 0
Then just open up the brackets and simplify further.
sqrt(3x + 7) + sqrt(3x - 7)= 0
Nothing to special really happened there, just removed the brackets. Now you move one of the radicals to the other side so you can square the whole equation.
sqrt(3x + 7) = - sqrt(3x - 7)
Then go ahead and square both sides to remove the radical.
3x + 7 = 3x - 7
Now if you kept trying to isolate x, you find that both sides will just cancel each other out and you are left with,
7 = -7
Since that statement isn't true your answer will be that there is no solution to this equation.
x ∈ Ø
$92,28-$18=$74.28
$74.28<span>÷$0.06= 1238minutes</span>
If x is number of hours worked and y total pay, in y=9.5x, 9.5 is hourly rate.
Isa's hourly rate is 9.5$.
From the table: for 2 hours worked Jack earns 16.5$, for 5 hours worked 41.25$ and for 8 66$. Jack's hourly rate is 16.5$/2=8.25$.
9.5/8.25=1.15
Isa earns 9.5$ per hour and Jack earns 8.25$ per hour. Isa's hourly rate of pay is 1.15 Jack's hourly rate of pay.