Hello, I need help please with this question if anyone could help me!

Yoooooooooooooooooooooooooo can you please help me!?

klemol [59]

She has totally 10+5+3=18 marbles, so the probability is 10/18=5/9. The complement is (9-5)/9=4/9.

Hope this helped!

4 0

2 years ago

NO websites or fake links, inappropriate comments, wired, or not helping comments our your reported and show me your steps to ea

Musya8 [376]

I couldn’t put the answer here here’s a link! Good luck!

Iwaskidding.lol

1. Is 2(1/3)
2. Is 4(1/5)
3. Is 3(1/9)
4. Is 4(1/2)
5. Is 3(1/12)

This is just an educated guess so please think about my answers carefully. Have a good day.

8 0

2 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

A weight clinic recorded the weight lost (in pounds) by each client of a weight control clinic during the last year, and got the

borishaifa [10]

Answer:

Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

31 - 40 ___________3 _________ 14

41 - 50 ___________ 1 _________ 15

51 - 60 ___________ 1 _________ 16

Step-by-step explanation:

Given the data :

35, 26, 31, 17, 46, 30, 28, 21, 26, 34, 15, 27, 7, 18, 16, 57

Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

31 - 40 ___________3 _________ 14

41 - 50 ___________ 1 _________ 15

51 - 60 ___________ 1 _________ 16

The next to highest frequency group has a frequency of 4 and the highest frequency of 6

Total frequency, n = (1 + 4 + 6 + 3 + 1 + 1) = 16

5 0

3 years ago

Consider the half of the sphere of radius 2 centered at the origin which lies above the xy-plane (a hemisphere). Suppose the den

Harlamova29_29 [7]

Answer:

25.133 units

Step-by-step explanation:

Since the density ρ = r, our mass is

m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So

m =∫∫[∫r³sinθdΦ]drdθ

m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π

m = ∫2πr³∫sinθdθ]dr

m = 2π∫r³dr ∫sinθdθ = 1

m = 2π × 4 ∫r³dr = 4

m = 8π units

m = 25.133 units

5 0

2 years ago