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mr Goodwill [35]
2 years ago
10

A yogurt shop offers 6 different flavors of frozen yogurt and 12 different toppings. How many choices are possible for a single

serving of frozen yogurt with one topping?
Mathematics
1 answer:
Goryan [66]2 years ago
7 0

Answer:

#3) 72; #4) 40,320; #5) 90.

Step-by-step explanation:

#3) The number of possible choices are found by multiplying the choices of flavors and the choices of toppings:

6*12=72.

#4) The ordering of 8 cards is a permutation, given by 8!=40,320.

#5) This is a permutation of 10 objects taken 2 at a time:

P(10,2) = 10!/(10-2)!=10!/8!=90.

P.S I had the same question once.

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Alex787 [66]

Answer:

P = \frac{132000}{12}=11000

So then P =11000 is the minimum that the least populated district could have.

Step-by-step explanation:

We have a big total of N = 132000 for the population.

And we know that we divide this population into 11 districts

And we have this info given "no district is to have a population that is more than 10 percent greater than the population of any other district"

Let's assume that P represent our minimum value for a district in the population. The range of possible values for the population of each district would be between P and 1.1 P

The interest on this case is find the minimum value for P and in order to do this we can assume that 1 district present the minimum and the other 10 the maximum value 1.1P in order to find which value of P satisfy this condition, and we have this:

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2 years ago
Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth Maclaurin polynomials for the function in
Zielflug [23.3K]

Answer:

Σ(-1)^kx^k for k = 0 to n

Step-by-step explanation:

The nth Maclaurin polynomials for f to be

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To calculate with a coefficient of 1

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Hence, the nth Maclaurin polynomials is

1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n

= Σ(-1)^kx^k for k = 0 to n

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