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2 years ago

The number of dairy cows in a farm can vary from month to month. The amount of milk

Mathematics

2 answers:

sattari [20]2 years ago

6 0

Answer:

G is dependent

N is independent

Effectus [21]2 years ago

5 0

Answer:

G is dependant d is independant

Step-by-step explanation:

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What is the probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from a

Drupady [299]

Answer:

P=0.00564

Step-by-step explanation:

From Exercise we have 52 cards.

We calculate the number of combinations to draw 5 cards from a deck of 52 cards. We get

{52}_C_{5}=\frac{52!}{5!(52-5)!}=2598960

We now count the number of favorable combinations:

{13}_C_{1} · {48}_C_{2}= 13 · \frac{48!}{2!(48-2)!}=14664

Therefore, the probabilitiy is

14664/2598960=0.00564

P=0.00564

6 0

3 years ago

a dental company has rolls of floss that are 1872 in long they want to put the distance in the yard on their label what is the d

statuscvo [17]

Copy and paste ur answer onto internet it will tell u

4 0

3 years ago

Please help !!! I’m stuck

Kruka [31]

Answer:

2/3 ≈ 0.6667

Step-by-step explanation:

apply soh cah toa

sine=opposite/hypo

sine=4/6

sine=2/3

4 0

2 years ago

Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition

charle [14.2K]

Answer:

The probability that there will be a total of 7 defects on four units is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let <em>X</em> = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, $(2\times4)=8$ .

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.

8 0

3 years ago

Need some help with this math question, can someone help me out pls

Sergeeva-Olga [200]

Okay I'm not too sure on this but I believe it is the third one.

I hope this helps!!

3 0

3 years ago