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3 years ago

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There are 7 green marbles, 8 blue marbles, and 4 yellow marbles. What is the chance, as a fraction, of drawing a blue marble, an

d without replacing it, drawing a green marble next?

2 answers:

alexandr1967 [171]3 years ago

4 0

1/19 and then to draw a green marble would have to be 1/18 so I hope that helps

luda_lava [24]3 years ago

3 0

There are 19 marbles total, when you add up all the colors. There are 8 blue marbles, so that's 8 out of 19, AKA 8/19. If you draw a blue marble and don't replace it, that's one less blue marble, and one less marble in general, AKA 7/18. So, now there are 18 marbles since you have drawn the blue one and haven't replaced it. There are 7 green marbles, and the total of marbles is 18 now. So, the chances of drawing a green marble is 7/18!

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Reyansh filled a 500-milliliter container with clean water. He then drank 35 percent of the water in the container. How many mil

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<span>He then drank 35 percent of the water in the container
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answer
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3 years ago

Read 2 more answers

Tickets for a certain show cost $17, $21, or, for VIP seats, $40. If ten times as many $17 tickets were sold as VIP tic

lawyer [7]

Answer:

Tickets sold:

VIP $=126$

$17 tickets $=1,260$

$21 tickets $=9\cdot 126+57=1,191$

Step-by-step explanation:

Let x be the number of VIP tickets.

If ten times as many $17 tickets were sold as VIP tickets, then the number of $17 tickets is $10x.$

If the number of $17 tickets sold was 57 more than the sum of the number of $21 tickets and VIP tickets, then $10x+57=x+y$ and the number y of $21 tickets is $9x+57.$

Amounts earned:

VIP tickets $=\$40x$

$17 tickets $=\$17\cdot 10x=\$170x$

$21 tickets $=\$21\cdot (9x+57)=\$(189x+1,197)$

Total $=\$(40x+170x+189x+1,197)=\$(399x+1,197)$

The sales of all three kinds of tickets would total $51,471, so

$399x+1,197=51,471\\ \\399x=51,471-1,197\\ \\399x=50,274\\ \\x=126$

Tickets sold:

VIP $=126$

$17 tickets $=1,260$

$21 tickets $=9\cdot 126+57=1,191$

8 0

3 years ago

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tamaranim1 [39]

Answer:

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Step-by-step explanation:

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3 years ago

Model the pair of situations with exponential functions f and g. Find the approximate value of x that makes f(x)= g(x).

Law Incorporation [45]

I like the cat on your pfp LOL sorry i couldn’t help tho

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3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago