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jeyben [28]
2 years ago
11

A bag has 10 marbles and 4 are black. Joseph picks 2 marbles without replacing the first. What is the probability that both are

black?
Mathematics
1 answer:
Crank2 years ago
6 0
If there are 10 marbles in total then the probability is 2/10
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In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
expeople1 [14]

The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

6 0
2 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, P(A)=0.95^{10}=0.5987

We know that the sum of probability of an event and its complement is 1.

So, P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

6 0
2 years ago
Read 2 more answers
If the instructions for a problem ask you to use the smallest possible domain to completely graph two periods of y = 5 + 3 cos 2
erica [24]
Hello,

y=5+3*cos (2(x-π/3))

The function is periodic with periode=2π.


-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8


8 0
2 years ago
You are 330 miles from home and you are driving toward home at an average of 55 mph. Write an equation to represent the situatio
Mars2501 [29]
Okay, I just know the answer not the range:

330 divided by 55 is 6. It would take 6 hours.
3 0
2 years ago
Read 2 more answers
The experimental probability that Kevin will catch a fly ball is equal to 7/8. About what percent of the time will Kevin catch a
Paladinen [302]

Answer:

87.5% percent chance.

Trust me it's right.

6 0
2 years ago
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