From the graph it seems to be around the area of 10%.
So,
18,000 * 0.10 = $1,800
Answer:
a₁ = 10
Step-by-step explanation:
The common difference d is calculated as
d = a₃ - a₂ = - 20 - (- 5) = - 20 + 5 = - 15
Then
a₂ - a₁ = d , that is
- 5 - a₁ = - 15 ( add 5 to both sides )
- a₁ = - 10 ( multiply both sides by - 1 )
a₁ = 10
(x-3)^2 + (y+5)^2 = 25
remove the ^2's from the left side:
(x-3) +(y+5) = sqrt(25)
(x-3) +(y+5) = 5
radius = 5
now solve for zero 's on the left side:
x-3 = 0 x = 3 (3-3=0)
y+5 = 0 y = -5 (-5 +5=0)
so center = (3,-5) and radius is 5
Answer is last one.
Answer:
CI=[0.8592,0.9402]
Yes, Method appears to be effective.
Step-by-step explanation:
-We first calculate the proportion of girls born:
Since np
, we assume normal distribution and calculate the 99% confidence interval as below:
![CI=\hat p\pm z\sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\=0.9\pm2.576\sqrt{\frac{0.9\times 0.1}{370}}\\\\=0.9\pm 0.0402\\\\={0.8598, \ 0.9402]](https://tex.z-dn.net/?f=CI%3D%5Chat%20p%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D%5C%5C%5C%5C%3D0.9%5Cpm2.576%5Csqrt%7B%5Cfrac%7B0.9%5Ctimes%200.1%7D%7B370%7D%7D%5C%5C%5C%5C%3D0.9%5Cpm%200.0402%5C%5C%5C%5C%3D%7B0.8598%2C%20%5C%200.9402%5D)
Hence, the confidence interval is {0.8598, 0.9402]
-The probability of giving birth to a girl is 0.5 which is less than the lower boundary of the confidence interval, it can be concluded that the method appears to be effective.
