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skad [1K]
2 years ago
8

Describe the difference between the value of x in a binomial distribution and in a geometric distribution.

Mathematics
1 answer:
Ganezh [65]2 years ago
8 0

Answer:

See below

Step-by-step explanation:

A binomial distribution has the value of x representing the number of successes in n​ trials, while a geometric distribution has the value of x representing the first trial that results in a success.

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There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun one time, what is the probability that the
NeX [460]

The probability that the result that is gotten will be a multiple of 3 and 2 will be 2/15.

<h3>How to calculate the probability</h3>

From the information, the spinner has 15 equal areas, numbered 1 through 15.

The multiple of 3 and a multiple of 2 among the numbers will be 6 and 15. Therefore, the probability will be two out of fifteen. This will be 2/15.

Learn more about multiples on:

brainly.com/question/1067440

8 0
2 years ago
A fraction f, multiplied by 5
prohojiy [21]

Answer:

f=1/40

Step-by-step explanation:

5f=1/8

f=(1/8)/5

f=(1/8)(1/5)

f=1/40

5 0
3 years ago
What is 27 over 40 as a decimal
baherus [9]
27 divided by 40 is equal to 0.675.
5 0
3 years ago
when the smallerof two consecutive integers is added to two times the larger,the result is 17.find the integers
Gnesinka [82]
Formula:

x + 2(x+1) = 17
x + 2x + 2 = 17
3x + 2 = 17
3x = 15
x = 5

So, the answer is 5 and 6

Hope this helped!
7 0
3 years ago
Can anyone tell me why by direct substitution of x, the equation (circled ones) equals to the indeterminate form, 0/0? When you
Katena32 [7]

Answer:

See explanation and hopefully it answers your question.

Basically because the expression has a hole at x=3.

Step-by-step explanation:

Let h(x)=( x^2-k ) / ( hx-15 )

This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.

Solving for x in that equation:

Adding 15 on both sides:

hx=15

Dividing both sides by h:

x=15/h

For it be a hole, you also must have the numerator is zero at x=15/h.

x^2-k=0 at x=15/h gives:

(15/h)^2-k=0

225/h^2-k=0

k=225/h^2

So if we wanted to evaluate the following limit:

Lim x->15/h ( x^2-k ) / ( hx-15 )

Or

Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.

We were ask to evaluate

Lim x->3 ( x^2-k ) / ( hx-15 )

Comparing the two limits h=5 and k=225/h^2=225/25=9.

3 0
2 years ago
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