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pychu [463]
3 years ago
13

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?
Mathematics
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
so 65% left in in the container

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answer
</span>325 milliliter of water left in the container
3 0
3 years ago
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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
ASAP<br> -WILL MARK BRIANIST
tamaranim1 [39]

Answer:

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Step-by-step explanation:

im not sure but the question asks about types of bread, since none are listed, i am assuming they all have a different price so the $1 bread $2 bread $3 bread and $5 bread... sorry i couldn't help better!

6 0
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]
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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
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