You most likely only need the first 3 digits after the decimal point but if you needed more i added them! hope this helps
A high school statistics class wants to estimate the average number of chocolate chips in a generic brand of chocolate chip cook
1 answer:
Answer:
Correct option:
We are 95% certain that the confidence interval of 18.6 to 21.3 includes the true average number of chocolate chips per cookie.
Step-by-step explanation:
The general formula for the (1 - <em>α</em>)% confidence interval for population<em> </em>mean is:
Here:
= sample mean
CV = critical value
= standard error of mean.
The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.
Or, the (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.
The 95% confidence interval for the mean number of chocolate chips per cookie is (18.6, 21.3).
This 95% confidence interval implies that there is a 0.95 probability that the true mean number of chocolate chips per cookie is between 18.6 and 21.3.
Thus, the correct option is:
"We are 95% certain that the confidence interval of 18.6 to 21.3 includes the true average number of chocolate chips per cookie. "
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Answer:
Volume of the frustum = ⅓πh(4R² - r²)
Step-by-step explanation:
We are to determine the volume of the frustum.
Find attached the diagram obtained from the given information.
Let height of small cone = h
height of the large cone = H
The height of a small cone is a quarter of the height of the large cone:
h = ¼×H
H = 4h
Volume of the frustum = volume of the large cone - volume of small cone
volume of the large cone = ⅓πR²H
= ⅓πR²(4h) = 4/3 ×π×R²h
volume of small cone = ⅓πr²h
Volume of the frustum = 4/3 ×π×R²h - ⅓πr²h
Volume of the frustum = ⅓(4π×R²h - πr²h)
Volume of the frustum = ⅓πh(4R² - r²)
Answer:
4 liters of 60% solution; 2 liters of 30% solution
Step-by-step explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
_____
If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.
