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3 years ago

How many ways can Gary, Cayla, and

Mathematics

1 answer:

Kryger [21]3 years ago

4 0

Answer:

Step-by-step explanation:

G,C,P

C,P,G

P,G,C

G,P,C

P,C,G

C,G,P

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2. Which interval notation represents the set of all real

anygoal [31]

B. (2,20] is the correct answer

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2 years ago

14. Set A contains all integers from 50 to 100, inclusive, and Set B contains all integers from 69 to 138, exclusive. How many i

tamaranim1 [39]

Answer:

Step-by-step explanation:

We have to make the intersection between these sets, and we see that they intersect in the range: [70,100], and d(70,100) = 30, that is the answer.

5 0

2 years ago

Based on a study of population projections for 2000 to 2050, the projected population of a group of people (in millions) can b

Olegator [25]

Answer:

(a) 0.107 million per year

(b) 0.114 million per year

Step-by-step explanation:

$A(t) = 11.19(1.009)^t$

(a) The average rate of change between 2000 and 2014 is determined by dividing the difference in the populations in the two years by the number of years. In the year 2000, $t=0$ and in 2014, $t=14$ . Mathematically,

$\text{Rate}=\dfrac{A(2014) - A(2000)}{2014-2000}=\dfrac{11.19(1.009)^14-11.19(1.009)^0} {14}$

$A(0)=\dfrac{12.69-11.19}{14}=\dfrac{1.5}{14}=0.107$

(b) The instantaneous rate of change is determined by finding the differential derivative at that year.

The result of differentiating functions of the firm $y=a^x$ (where $a$ is a constant) is $\dfrac{dy}{dx}=a^x\ln a$ . Let's use in this in finding the derivative of $A(t)$ .

$A\prime(t) = \dfrac{A}{t}=11.19\cdot1.009^t\ln1.009$

In the year 2014, $t=14$ .

$A\prime(14) =11.19\cdot1.009^14\ln1.009=0.114$

6 0

2 years ago

Taxes on 100 dlls??

dezoksy [38]

The sales tax is 6.25% from $100

8 0

2 years ago

Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair whic

docker41 [41]

Answer:

Let f_n be the number of rabbit pairs at the beginning of each month. We start with one pair, that is f_1 = 1. After one month the rabbits still do not produce a new pair, which means f_2 = 1. After two months a new born pair appears, that is f_3 = 2, and so on. Let now n $\geq$ 3 be any natural number. We have that f_n is equal to the previous amount of pairs f_n-1 plus the amount of new born pairs. The last amount is f_n-2, since any two month younger pair produced its first baby pair. Finally we have

f_1 = f_2 = 1,f_n = f_n-1 + f_n-2 for any natural n $\geq$ 3.

5 0

3 years ago