Answer: $10.20
Step-by-step explanation:
Let's first collect the information from the word problem.
<h3><em>We are told:</em></h3>
- Bought 15 pies and 7 cake slices
- Total cost was $55.22.
- Pies cost
as much as a slice of cake
<h3><em>We need to solve for:</em></h3>
- Total cost of 1 pie and 2 cake slices
1. Create an equation/expression:
To solve the problem, we need to find the price of a pie and cake slice.
Using the information gathered, the equation is:
<em>x = the cost of the slice of cake.</em>
<em />
2. Simplify/Solve
- Great! we have the price of a slice of cake. ($4.25)
Remember that the pie is of a slice of cake, so now find the price of the pies.
Pie = 
Pie = $1.70
- Awesome! We know a slice of cake costs $4.25 and a slice of pie costs $1.70.
3. Check your work!
<em>Check to make sure the math is done correctly!</em>
Good! It is verified that our calculations are correct.
4. Solve!
1 pie and 2 cakes cost: 
- $10.20 is our final answer!
- Hope this helped!~~ -Astro
Start with #47. To find the critical values, you must differentiate this function. x times (4-x)^3 is a product, so use the product rule. The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x]
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of f '(x) = (4-x)^2 [-3x+4-x]. (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1. Thus, the "cv" are {4,1}.
Evaluate the given function at x: {4,1}. For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8. Thus, one of the extreme values is (1,8).
P(white) = 27/100 = 270/1000
Therefore, in 1,000 pulls, we eaxpect 270 white beads,
Answer: Paired t interval for μdiff
Step-by-step explanation:
The confidence interval suitable for analysing the data described in the scenario above is the paired t test which is employed when there are two different sets of measured reading or observation for each subject. In the scenario above, the subjects are of of the different planters employed whereby the pair of measurement are the weight of tomatoes obtained for each of the two different fertilizer types applied to each of the two plants in a planter. With these data, we can measure the difference in the effectiveness of each of the store brand and homegrown compost.
