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ziro4ka [17]
3 years ago
8

You randomly draw a marble from a bag of marbles that contains 8 88 blue marbles, 5 55 green marbles, and 8 88 red marbles. What

is P(draw a blue or red marble ) P(draw a blue or red marble)start text, P, left parenthesis, d, r, a, w, space, a, space, b, l, u, e, space, o, r, space, r, e, d, space, m, a, r, b, l, e, end text, right parenthesis? If necessary, round your answer to 2 22 decimal places.
Mathematics
2 answers:
Artyom0805 [142]3 years ago
5 0

Answer:

0.25

Step-by-step explanation:

3 favorable outcomes

 

3+4+5=123 + 4 + 5 = 12

3+4+5=12

3, plus, 4, plus, 5, equals, 12

total marbles

P(draw a blue marble)=

12

3

​

=0.2

Fantom [35]3 years ago
4 0

Answer:

7%

Step-by-step explanation:

add all numbers up get 21 then divide by 3 and get 7

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2 years ago
What is the product of (X4/3)(X2/3) ?
katrin [286]
Hi, the answer to this would be x6/9. I'm assuming the x4/3 and x2/3 are fractions and the x's aren't exponents. Now how I got x6/9 is shown here.

1st Step: Started off by regrouping the terms
1/3x3 x^4x^2

2nd Step: we can easily simplify 3x3 to just 9. And now we're left with 1/9x^4x^2

3rd Step: Now we can simplify the 1/9 to just x^4x^2/9

4th Step: Now we can use the product rule which is simple. So We add the exponents and simplify it to just one exponent. So x4+2=6 that simplifies to just x^6.

Final Answer: x^6/9.
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3 years ago
Berry Delicious is a popular shop that sells chocolate-covered strawberries. Last year, the shop used 5,900 kilograms of strawbe
Ugo [173]
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3 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
3 years ago
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