Answer:
a) The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is 10.02.
Step-by-step explanation:
The t-distribution is used to solve this question:
a) n = 14
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 14 - 1 = 13
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7709
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) n = 28
27 df, T = 1.7033

The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is
44 df, T = 1.6802

The margin of error for a 90% confidence interval when n = 45 is 10.02.
Answer:
a
Step-by-step explanation:
y= 1/2 ^(x+4)
Low-birthweight i believe.
The first thing you must do is to calculate the weight of the gravel (w) that the truck holds (7.5 yards³). So, you have:
If the weight of 1 yard³ is 1.48 tons, then the weight of 7.5 yards³ is:
w=7.5x1.48 tons
w=11.1 tons
The problem says that the gravel will be placed in containers that each hold 3.7 tons of gravel,so, If they need 1 cotainer to place 3.7 tons, how many containers they need to place 11.1 tons?
1 container-------3.7 tons of gravel
x-------11.1 tons of gravel
x=(11.1x1)/3.7
x=11.1/3.7
x=3 containers
H<span>ow many containers of this size are needed to hold all the gravel from one truck?
</span>
The answer is: 3 containers.
Answer:
green
Step-by-step explanation:
the one with the most gumballs compared to the other colors is the most likely