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strojnjashka [21]
3 years ago
15

PLZZ HELP! Sara selects two cards at random from a standard deck of fifty-two cards. Which of the following could be used to cal

culate the probability that she will select two even-numbered cards if she does not replace the first card before selecting the second? Note: For this problem, face cards and aces are not numbered cards.
sixteen over fifty-two times sixteen over fifty-one

sixteen over fifty-two times fifteen over fifty-one

sixteen over fifty-two times fifteen over fifty-two

sixteen over fifty-two times sixteen over fifty-two
Mathematics
1 answer:
patriot [66]3 years ago
5 0

Answer:

91

Step-by-step explanation:

it takes to long to explain

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Solve and show how pls
irakobra [83]

Answer:

16

Step-by-step explanation:

since -4 is being squared, we can write it out as -4 * -4, to get us the answer 16. this answer is positive because a negative times a negative will cancel out and = a positive.

4 0
2 years ago
HURRYYY PLEASE THIS IS A TESTTT <br> Find the value of x.
o-na [289]

Answer:

130° is the answer to it

Step-by-step explanation:

180-50=130

6 0
3 years ago
Read 2 more answers
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}

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2 years ago
Plz help ill give you brainlist
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Bananas are on sale at the grocery store.
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Add them together thee we j get your answer
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