(25ab^3)^1/3 divided by (ab)^-1/2
1 answer:
You might be interested in
Answer:
98% confidence interval for the average age of all students is [24.302 , 25.698]
Step-by-step explanation:
We are given that a random sample of 36 students at a community college showed an average age of 25 years.
Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.
So, the pivotal quantity for 98% confidence interval for the average age is given by;
P.Q. = ~ N(0,1)
where, = sample average age = 25 years
= population standard deviation = 1.8 years
n = sample of students = 36
= population average age
So, 98% confidence interval for the average age, is ;
P(-2.3263 < N(0,1) < 2.3263) = 0.98
P(-2.3263 < < 2.3263) = 0.98
P( <
<
) = 0.98
P( <
<
) = 0.98
98% confidence interval for = [
,
]
= [ ,
]
= [24.302 , 25.698]
Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].
Answer:
H. 552
Step-by-step explanation:
Circle graph information:
- Milk and eggs = x%
- Fruits and vegetables = 38%
- Cereals = 10%
- Meat and fish = 10%
- Sugar = 8%
- Fats and oils = 4%
Total consumption = 1840 pounds per person
First, calculate the percentage of milk and eggs (the value of x).
Total percentages = 100%
⇒ x + 38 + 10 + 10 + 8 + 4 = 100
⇒ x + 70 = 100
⇒ x = 30
Therefore, 30% of food consumption is milk and eggs.
To calculate the number of pounds of milk and eggs the average American consumes each year, simply find 30% of the total consumption:
= 30% of 1840
= 30% × 1840
= 552
Therefore, the average American consumes 552 pounds of milk and eggs each year.