Answer:
The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
![n = 1600, \pi = 0.4](https://tex.z-dn.net/?f=n%20%3D%201600%2C%20%5Cpi%20%3D%200.4)
99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.3685](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.4%20-%202.575%5Csqrt%7B%5Cfrac%7B0.4%2A0.6%7D%7B1600%7D%7D%20%3D%200.3685)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.4315](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.4%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.4%2A0.6%7D%7B1600%7D%7D%20%3D%200.4315)
The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Simplifying
(0.75x + 6) + -1(2.5x + -1.9) = 0
Reorder the terms:
(6 + 0.75x) + -1(2.5x + -1.9) = 0
Remove parenthesis around (6 + 0.75x)
6 + 0.75x + -1(2.5x + -1.9) = 0
Reorder the terms:
6 + 0.75x + -1(-1.9 + 2.5x) = 0
6 + 0.75x + (-1.9 * -1 + 2.5x * -1) = 0
6 + 0.75x + (1.9 + -2.5x) = 0
Reorder the terms:
6 + 1.9 + 0.75x + -2.5x = 0
Combine like terms: 6 + 1.9 = 7.9
7.9 + 0.75x + -2.5x = 0
Combine like terms: 0.75x + -2.5x = -1.75x
7.9 + -1.75x = 0
Solving
7.9 + -1.75x = 0
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-7.9' to each side of the equation.
7.9 + -7.9 + -1.75x = 0 + -7.9
Combine like terms: 7.9 + -7.9 = 0.0
0.0 + -1.75x = 0 + -7.9
-1.75x = 0 + -7.9
Combine like terms: 0 + -7.9 = -7.9
-1.75x = -7.9
Divide each side by '-1.75'.
x = 4.514285714
Simplifying
x = 4.514285714
Answer:
a-am i really cute
Step-by-step explanation:
Answer:
New cars sold = 20,
Used cars sold = 25
Cars serviced = 290
Step-by-step explanation:
In a month they sold fifteen less used cars than twice the number of new cars.
Lets say 'x' number of new cars were sold, then:
The number of used car sold is:
![2\times x-15=2x-15](https://tex.z-dn.net/?f=2%5Ctimes%20x-15%3D2x-15)
In the same month they serviced fifty more than twelve times the number of new cars sold. So the number of cars serviced is:
![12\times x+50=12x+50](https://tex.z-dn.net/?f=12%5Ctimes%20x%2B50%3D12x%2B50)
Altogether they sold or serviced 335 cars, so the sum of all the cars sold and serviced should be 335:
![x+(2x-15)+(12x+15)=335](https://tex.z-dn.net/?f=x%2B%282x-15%29%2B%2812x%2B15%29%3D335)
Solving for 'x' we get:
![x+2x+12x-15+50=335](https://tex.z-dn.net/?f=x%2B2x%2B12x-15%2B50%3D335)
![15x+35=335](https://tex.z-dn.net/?f=15x%2B35%3D335)
![15x=335-35=300](https://tex.z-dn.net/?f=15x%3D335-35%3D300)
![x=\frac{300}{15}=20](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B300%7D%7B15%7D%3D20)
Therefore, the number of new cars sold in that month is 20.
Number of used car sold ![=2x-15=2\times 20-15=40-15=25](https://tex.z-dn.net/?f=%3D2x-15%3D2%5Ctimes%2020-15%3D40-15%3D25)
Number of cars serviced ![=12\times x+50=12\times20+50=240+50=290](https://tex.z-dn.net/?f=%3D12%5Ctimes%20x%2B50%3D12%5Ctimes20%2B50%3D240%2B50%3D290)
We can also cross check by summing them up:
![290+25+20=335](https://tex.z-dn.net/?f=290%2B25%2B20%3D335)