Answer:
a) X[bar]₁= 1839.20 cal
b) X[bar]₂= 1779.07 cal
c) S₁= 386.35 cal
Step-by-step explanation:
Hello!
You have two independent samples,
Sample 1: n₁= 15 children that did not eat fast food.
Sample 2: n₂= 15 children that ate fast food.
The study variables are:
X₁: Calorie consumption of a kid that does not eat fast food in one day.
X₂: Calorie consumprion of a kid that eats fast food in one day.
a)
The point estimate of the population mean is the sample mean
X[bar]₁= (∑X₁/n₁) = (27588/15)= 1839.20 cal
b)
X[bar]₂= (∑X₂/n₂)= (26686/15)= 1779.07 cal
c)
To calculate the sample standard deiation, you have to calculate the sample variance first:
S₁²= 
[∑X₁² - (( ∑X₁)²/n₁)]= 
= 149263.4571 cal²
S₁= 386.35 cal
I hope it helps!
Answer:
three boys shared D 10,500.00 in the ratio 6:7:8.find the largest share.
Answer: a - 4.512 hours
b - 1.94 hours
Step-by-step explanation:
Given,
a) A(t) = 10 (0.7)^t
To determine when 2mg is left in the body
We would have,
A(t) = 2, therefore
2 = 10(0.7)^t
0.7^t =2÷10
0.7^t = 0.2
Take the log of both sides,
Log (0.7)^t = log 0.2
t log 0.7 = log 0.2
t = log 0.2/ 0.7
t = 4.512 hours
Thus it will take 4.512 hours for 2mg to be left in the body.
b) Half life
Let A(t) = 1/2 A(0)
Thus,
1/2 A(0) = A(0)0.7^t
Divide both sides by A(0)
1/2 = 0.7^t
0.7^t = 0.5
Take log of both sides
Log 0.7^t = log 0.5
t log 0.7 = log 0.5
t = log 0.5/log 0.7
t = 1.94 hours
Therefore, the half life of the drug is 1.94 hours
Is their any multiple questions
Answer:
Thank you so much. She has even deleted one of my questions before. I'll take care and beware.
