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2 years ago

can someone help me in solving this with explanation please.. i really need it and i will be so thankful

Mathematics

1 answer:

arsen [322]2 years ago

8 0

Answer:

The correct answer is C: 5 root 2

Step-by-step explanation:

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Which statement best describes the use of a simulation to predict the probability that two randomly chosen people will both have

laila [671]

Randomly generate an integer from 1 to 7 two times, and the probability is 1/7 ^2

This is the statement that best describes the use of a simulation to predict the probability that two randomly chosen people will both have their birthdays on a Monday.

There are 7 days in a week, so there are 7 choices but only 1 Monday. So, 1/7 is the probability that a person's birthday falls on a Monday.

1st person asked will have 1/7 probability.
2nd person asked will also have 1/7 probability

So, (1/7)² is the probability that both persons will have their birthdays on a Monday.

8 0

2 years ago

Read 2 more answers

3.x4 - 2x2 + 15.x2 Which product Is the factored form of the polynomial

Dmitry_Shevchenko [17]

Answer:

3*4=12

12-2=10

10*2=20

20+15=35

35*2=70

Step-by-step explanation:

4 0

2 years ago

Determine if each function is linear or nonlinear.

Akimi4 [234]

Answer:

Linear: y = x, y = x/2 - 3, 3x + 2 = 12

Nonlinear: y = 6/x - 2, y = 3x^3 + 5

Step-by-step explanation:

Linear functions form straight lines while nonlinear functions do not.

3 0

2 years ago

Is it A. B. C. or D.

podryga [215]

Your answer would be b i'm pretty sure

5 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

2 years ago