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docker41 [41]
3 years ago
7

suppose the monkey is typing using only the 26 letter Keys what is the probability that the monkey will type cat

Mathematics
1 answer:
Alex787 [66]3 years ago
3 0
Every time the monkey types 3 letters, the probability that they will be C-A-T is
(1/26)-cubed = about 0.0057 % .
You might be interested in
14% increase of 43200
Tomtit [17]
43,200 is 100%, plus 14%. You are looking for 114% of 43,200.

To find this, simply multiply 43,200*1.14, which equals 49,248.
7 0
3 years ago
Would anyone mind showing me how to do these questions? Thanks to anyone who helps out :)
cupoosta [38]
Question 1 is 40 
question 2 is 96
7 0
4 years ago
Read 2 more answers
Alejandro surveyed his classmates to determine who has ever gone surfing and who has ever gone snowboarding. Let A be the event
victus00 [196]

Answer:

{A and B are independent events}, P(A|B)=P(A)=0.16

Step-by-step explanation:

First of all we need to know when does two events become independent:

For the two events to be independent,  P(A|B)=P(A) that is if condition on one does not effect the probability of other event.

Here, in our case the only option that satisfies the condition for the events to be independent is  P(A|B)=P(A)=0.16.. Rest are not in accordance with the definition of independent events.

3 0
3 years ago
Read 2 more answers
Let f be a differentiable function such that f(1)=π and f'(x)=√x^3+6. what is the value of f(5)?
natali 33 [55]

The value of f(5) is 49.1

Step-by-step explanation:

To find f(x) from f'(x) use the integration

f(x) = ∫ f'(x)

1. Find The integration of f'(x) with the constant term

2. Substitute x by 1 and f(x) by π to find the constant term

3. Write the differential function f(x) and substitute x by 5 to find f(5)

∵ f'(x) = \sqrt{x^{3}} + 6

- Change the root to fraction power

∵ \sqrt{x^{3}} = x^{\frac{3}{2}}

∴ f'(x) = x^{\frac{3}{2}} + 6

∴ f(x) = ∫ x^{\frac{3}{2}} + 6

- In integration add the power by 1 and divide the coefficient by the

 new power and insert x with the constant term

∴ f(x) = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 6x + c

- c is the constant of integration

∵ \frac{x^{\frac{5}{2}}}{\frac{5}{2}}=\frac{2}{5}x^{\frac{5}{2}}

∴ f(x) = \frac{2}{5} x^{\frac{5}{2}} + 6x + c

- To find c substitute x by 1 and f(x) by π

∴ π = \frac{2}{5} (1)^{\frac{5}{2}} + 6(1) + c

∴ π = \frac{2}{5} + 6 + c

∴ π = 6.4 + c

- Subtract 6.4 from both sides

∴ c = - 3.2584

∴ f(x) = \frac{2}{5} x^{\frac{5}{2}} + 6x - 3.2584

To find f(5) Substitute x by 5

∵ x = 5

∴ f(5) = \frac{2}{5} (5)^{\frac{5}{2}} + 6(5) - 3.2584

∴ f(5) = 49.1

The value of f(5) is 49.1

Learn more:

You can learn more about differentiation in brainly.com/question/4279146

#LearnwithBrainly

4 0
3 years ago
A particular brand of dishwasher soap is sold in three sizes: 25 oz, 45 oz, and 60 oz. Twenty percent of all purchasers select a
Verdich [7]

Answer:

(a) <u>Sampling distribution </u>

P(25) = 0,04

P(35) = 0.1 + 0.1 = 0,2

P(42,5) = 0.06 + 0.06 = 0,12

P(45) = 0,25

P(52,5) = 0.15 + 0.15 = 0,3

P(60) = 0,09

(b) E(X) = 45.5 oz

(c) E(X) = μ

Step-by-step explanation:

The variable we want to compute is

X=(X1+X2)/2

For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.

(a) <u>Sampling distribution </u>

Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):

P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04

P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2

P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12

P(45) = p45*p45 = 0.5 * 0.5 = 0,25

P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3

P(60) = p60*p60 = 0.3*0.3 = 0,09

(b) Using the sample distribution, E(X) can be expressed as:

E(X)=\sum_{i=1}^{6}P_{i}*X_{i}\\E(X)=0.04*25+0.2*35+0.12*42.5+0.3*52.5+0.09*60 = 45.5

The value of E(X) is 45.5 oz.

(c) The value of μ can be calculated as

\mu=\sum_{i=1}^{3}P_{i}*X_{i}\\\mu=0.2*25+0.5*45+0.3*60=45.5

We can conclude that E(X)=μ

We could have arrived to this conclusion by applying

E(X)=E((X1+X2)/2)=E(X1)/2+E(X2)/2\\\\\mu = E(X1)=E(X2)\\\\E(X)=\mu /2+ \mu /2 = \mu

4 0
4 years ago
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