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galben [10]
3 years ago
13

A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops

, what is the probability that the cricket has returned to the leaf where it started?
I NEED HELP ASAP PLS
Mathematics
1 answer:
Agata [3.3K]3 years ago
3 0

The probability that the cricket has returned to the leaf where it started is

  • \frac{1}{4}

After its 1st jump since it has the option to jump back to any leaf and it is the case of independent events. So probability will remain\frac{1}{4}.

<h3>What is probability?</h3>

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true.

For more information on probability, visit

brainly.com/question/24756209

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it answer is c ,d and f

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Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C
netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}</span></span>


Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.


|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5

|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6

|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4

|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \  \sqrt{25}=5

|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4


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3 0
3 years ago
Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang
PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence  f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence  f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

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substitute x = 0 into the function

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y = 0^2+2

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Answer:

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