Answer: 0.03855
Step-by-step explanation:
Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.
Its maximum safe load is 10000 pounds.
Let X denotes the weight of 50 people.
As per given ,
Population mean weight of 50 people =
Standard deviation of 50 people 
Then , the probability its maximum safe load will be exceeded =
![P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855](https://tex.z-dn.net/?f=P%28X%3E10000%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B10000-9500%7D%7B282.84%7D%29%5C%5C%5C%5C%3DP%28z%3E1.7671-8%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1.7678%29%5C%20%5C%20%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%3Ez%29%3DP%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3D1-0.96145%5C%20%5C%20%5C%20%5B%5Ctext%7B%20By%20p-value%20of%20table%7D%5D%5C%5C%5C%5C%3D0.03855)
Thus , the probability its maximum safe load will be exceeded = 0.03855
<h2>
Answer with explanation:</h2>
Let p be the population proportion of parents who had children in grades K-12 were satisfied with the quality of education the students receive.
Given : Several years ago, 39% of parents who had children in grades K-12 were satisfied with the quality of education the students receive.
Set hypothesis to test :

Sample size : n= 1055
Sample proportion : 
Critical value for 95% confidence : 
Confidence interval : 

Since , Confidence interval does not contain 0.39.
It means we reject the null hypothesis.
We conclude that 95% confidence interval represents evidence that parents' attitudes toward the quality of education have changed.
(4,4) are the coordinates of C prime
Answer:

Just like last time, you can convert them to their decimal forms, subtract, and convert back!
Hope this helps!
Brainliest please!
5 mi = 26400 ft.
5 mi is bigger by 100 ft.
~Evie