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NikAS [45]
3 years ago
14

Type the correct answer in each box. If necessary, use / for the fraction bar. A bag contains 5 blue marbles, 2 black marbles, a

nd 3 red marbles. A marble is randomly drawn from the bag. The probability of not drawing a black marble is . The probability of drawing a red marble is .
Mathematics
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

Step-by-step explanation:

Problem One

Blue =   5

Black  =2

Red    = 3

First of all there are 10 marbles, 2 of which are black.

That means that 8 others are not black

You can draw any one of the 8.

P(not black) = 8/10 = 4/5

Problem Two

There are 10 marbles in all

3 of them are red.

P(Red) = 3/10  

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If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the va
daser333 [38]

Answer:

a=2

Step-by-step explanation:

Well, first let us find product of the first 8 positive integers, which is 1*2*3*4*5*6*7*8=40320. It can also be written as 40320=2^{7}*3²*5*7. From the formula above, it can be extracted that if a=2, n can be any number from 3 to 7 (considering that a and n are not equal and greater than 1).

In (1) a^n=64 and in (2) n=6 so it can be written as 2^6=64 so a=2 and n=6. So the value of a is 2. a=2

3 0
3 years ago
A cylinder with a radius of 1 cm and a height of 21 cm has the same volume as a cone with a height of 7 cm. What is the radius o
Natasha_Volkova [10]

Answer:

<h2>A) 3 cm</h2>

Step-by-step explanation:

The formula of a volume of a cylinder:

V=\pi r^2H

r - radius

H - height

We have r = 1cm and H = 21cm. Substitute:

V=\pi(1^2)(21)=21\pi\ cm^3

The formula of a cone:

V=\dfrac{1}{3}\pi r^2H

r - radius

H - height

We have V = 21π cm³ and H = 7cm. Substitute:

\dfrac{1}{3}\pi(r^2)(7)=21\pi        <em>divide both sides by π</em>

\dfrac{1}{3}(7)(r^2)=21         <em>divide both sides by 7</em>

\dfrac{1}{3}r^2=3         <em>multiply both sides by 3</em>

r^2=9\to r=\sqrt9\\\\r=3\ cm

5 0
3 years ago
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10 choices for 1st slot
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3 years ago
How do I draw a model that shows 0.16
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3 years ago
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10 marbles are placed in a bag and two of them are randonly drawn.
Gelneren [198K]
The problem does not say anything about how many of the marbles are yellow but if you put the first one back in the bag then it is an independent probability and each one would be the same individual probability. Since the problem doesn't say and I don't know if you have it somewhere else, I'll do it as an example.

Let's say there were 6 of the 10 marbles were yellow. So the probability of the first one is 6/10 and so is the second one 6/10. To find the total probability you multiply them together 6/10*6/10=36/100=9/25

Hope that helps. If you can tell me how many yellow there are I will show the work for the specific problem.
4 0
3 years ago
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