Answer:
16 cups of cashews
12 cups of almonds
Step-by-step explanation:
Let's say x is the cups of cashews and y is the cups of almonds.
We can write two equations:
x / y = (1/3) / (1/4) = 4/3
x + y = 28
Solve the system of equations using substitution.
x = 4/3 y
4/3 y + y = 28
7/3 y = 28
y = 12
x = 16
Answer:
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
![\sigma = 5 .7](https://tex.z-dn.net/?f=%5Csigma%20%3D%205%20.7)
The standard deviation will remain unchanged.
Step-by-step explanation:
Given
![Data: 136, 129, 141, 139, 138, 127](https://tex.z-dn.net/?f=Data%3A%20136%2C%20129%2C%20141%2C%20139%2C%20138%2C%20127)
Solving (a): The range
This is calculated as:
![Range = Highest - Least](https://tex.z-dn.net/?f=Range%20%3D%20Highest%20-%20Least)
Where:
![Highest = 141; Least = 127](https://tex.z-dn.net/?f=Highest%20%3D%20141%3B%20Least%20%3D%20127)
So:
![Range=141-127](https://tex.z-dn.net/?f=Range%3D141-127)
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
Solving (b): The variance
First, we calculate the mean
![\bar x = \frac{1}{n} \sum x](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%5Csum%20x)
![\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%28136%2B%20129%2B%20141%2B%20139%2B%20138%2B%20127%29)
![\bar x = \frac{1}{6} *810](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%2A810)
![\bar x = 135](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20135)
The variance is calculated as:
![\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%28x%20-%20%5Cbar%20x%29%5E2)
So, we have:
![\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B6-1%7D%2A%5B%28136%20-%20135%29%5E2%20%2B%28129%20-%20135%29%5E2%20%2B%28141%20-%20135%29%5E2%20%2B%28139%20-%20135%29%5E2%20%2B%28138%20-%20135%29%5E2%20%2B%28127%20-%20135%29%5E2%5D)
![\sigma^2 =\frac{1}{5}*[162]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B5%7D%2A%5B162%5D)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
Solving (c): The standard deviation
This is calculated as:
![\sigma = \sqrt {\sigma^2 }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B%5Csigma%5E2%20%7D)
![\sigma = \sqrt {32.4}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B32.4%7D)
--- approximately
Solving (d): With the stated condition, the standard deviation will remain unchanged.
Answer:
112.5 mL
Step-by-step explanation:
Let x represent the amount of grape juice to add. Then the total amount of grape juice in the mix is ...
0.15(1800) +x = 0.20(1800+x)
270 +x = 360 + 0.2x . . . . . . eliminate parentheses
0.8x = 90 . . . . . . . . . . . . . . . subtract 270+0.2x
x = 90/0.8 = 112.5 . . . . . . . . divide by the coefficient of x
112 mL of grape juice should be added to make a mix that is 20% grape juice.
Answer:
90% confidence interval -> {0.4529, 0.5871}
Step-by-step explanation:
<u>Check conditions for a 1-proportion z-interval:</u>
np>10 -> 150(0.52)>10 -> 78>10 √
n(1-p)>10 -> 150(1-0.52)>10 -> 72>10 √
Random sample √
n>30 √
For a 90% confidence interval, the critical value is z=1.645
The formula for a confidence interval is:
CI = p ± z√[p(1-p)/n]
<u>Given:</u>
p = 78/150 = 0.52
n = 150
z = 1.645
Therefore, the 90% confidence interval is:
CI = 0.52 ± 1.645√[0.52(1-0.52)/150] = {0.4529, 0.5871}
Context: We are 90% confident that the true proportion of all voters who
plan to vote for the incumbent candidate is contained within the interval
Answer:
A.
Step-by-step explanation: