I’m 85% sure it’s the long line on the bottom if not then uhm I guess I’m wrong
Answer:
We know that:
She needs 9/10 grams of flour for the cake (this quantity does not make a lot of sense, maybe is written incorrectly, but is the only info we have, so let's solve the problem with this)
She only has 1/3 of the amount she needs.
How much flour does she have?
Well, she needs 9/10 grams, and she has one-third of that.
Then she has (1/3) times (9/10) grams, this is:
F = (1/3)*(9/10) grams
F = (1*9)/(3*10) grams
F = 9/30 grams
That is the amount of flour that she has.
Answer:
13 1/3 miles per hour
Step-by-step explanation:
Take the distance and divide by the time
1 1/3 ÷ 1/10
Change to an improper fraction
4/3 ÷ 1/10
Copy dot flip
4/3 * 10/1
40/3 miles per hour
changing to a mixed number
13 1/3 miles per hour
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.
