Answer:
The number of pounds of cashew is 11
Step-by-step explanation:
Let P represent the peanut
Let C represent the cashew nut.
From question, we were told that Elijah brought a total of 16 pounds of peanuts and cashew nuts. This can be written as:
P + C = 16 (1)
The total cost = $49.50
Cost per peanut = $2.75
Cost per cashew = $3.25
The above can be represented as:
49.50 = 2.75P + 3.25C. (2)
From equation 1,
P + C = 16
P = 16 — C
Substitute the value of P into equation 2:
49.50 = 2.75P + 3.25C.
49.50 = 2.75(16 — C) + 3.25C
49.50 = 44 — 2.75C + 3.25C
49.50 = 44 + 0.5C
Collect like terms
49.50 — 44 = 0.5C
5.5 = 0.5C
Divide both side by 0.5
C = 5.5/0.5
C = 11
Therefore, the number of pounds of cashew is 11
Answer:
The margin of error for this estimate is of 14.79 yards per game.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
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 = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 2.093
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.
This means that 
What is the margin of error for this estimate?
The margin of error for this estimate is of 14.79 yards per game.
Find the midpoint of (1,-4) and (7,1) to find the rest stop. To do this add the x values and divide by 2, then add the y values and divide by 2. (4,-1.5) is the midpoint.
To find the distance, use the distance formula then multiply the result by 5.38. The distance is 42 miles
Your answer will be 60%
Have a nice day
Answer:
a. [ 0.454,0.51]
b. 599.472 ~ 600
Step-by-step explanation:
a)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=410
Sample Size(n)=850
Sample proportion = x/n =0.482
Confidence Interval = [ 0.482 ±Z a/2 ( Sqrt ( 0.482*0.518) /850)]
= [ 0.482 - 1.645* Sqrt(0) , 0.482 + 1.65* Sqrt(0) ]
= [ 0.454,0.51]
b)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.482
ME = 0.04
n = ( 1.96 / 0.04 )^2 * 0.482*0.518
= 599.472 ~ 600
