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21 day ago

8

You are given a bag of coins with varying biases.the probability of heads is a random variable sample from uniform distribution

u[0,1].you draw a coin from these bag and toss it 100 times what is the probability of getting 100 heads stackover flow

1 answer:

Inessa [9K]21 day ago

4 0

Answer:

Step-by-step explanation:

Imagine having a collection of n biased coins, and you draw m<n of them without replacement, subsequently measuring each coin i for its parameter pi∈[0,1], indicating that each coin behaves as Bernoulli(pi). Now, I am curious to determine the most probable pm+1 for the next coin I choose. The only method I can think of is calculating the average of the parameters of the m coins sampled thus far, which can be expressed as: p^m+1=p1+…+pmm.

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The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [9117]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

7 0

1 month ago

Read 2 more answers

A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer

babunello [8426]

Answer:

The top surface area of the washer equals 160.14 square millimeters.

Step-by-step Explanation:

The washer's top surface forms an annulus, characterized by an outer radius of 10 mm and an inner radius of 7 mm (obtained since the hole's diameter is 14 mm and the radius is half the diameter).

Recall the formula for the area of an annulus:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r the inner radius.

Substituting the given values:

$R=10\ mm\\\\and\\\\r=7\ mm$

Thus, the calculation yields:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$

4 0

1 month ago

Read 2 more answers

Which statement is true about the polynomial 3j4k−2jk3+jk3−2j4k+jk3 after it has been fully simplified? in terms and degree

zzz [9093]

I just took this assessment and got it right; the answer is a term with a degree of 5. It’s option 4.

7 0

23 days ago

5 Show different ways to make 492,623.

zzz [9093]

Step-by-step explanation:

Begin with expressing 492,623 in standard form.

4 hundred thousands + 9 ten thousands + 2 thousands + 6 hundreds + 2 tens + 3 ones.

We can rephrase this in varied forms by shifting a digit to the next lower place value. For instance, shifting the 4 one place right results in 49 ten thousands:

49 ten thousands + 2 thousands + 6 hundreds + 2 tens + 3 ones.

Next, we can move 49 ten thousands one place right to express it as 492 thousands, and shift 6 hundreds right to yield 62 tens.

492 thousands + 62 tens + 3 ones.

Alternatively, we can phrase it as:

4926 hundreds + 23 ones.

6 0

26 days ago

A barbershop requires appointments for perms and hair cuts. Ten percent of those making appointments cancel or simply fail to sh

tester [8851]

Answer:

0.8937

Step-by-step explanation:

This involves binomial probability with n = 64, p = 0.10, and x = 3. This indicates a 10% chance of cancellations. To determine the likelihood of having more than 3 cancellations or no-shows, we calculate binompdf(64,0.10,0) + binompdf(64,0.10,1) + binompdf(64,0.10,2) + binompdf(64,0.10,3), then subtract that total from 1.000.

The result is: 0.0012 + 0.0084 + 0.0293 + 0.0674 = total = 0.1063

Thus, the target probability is 1.0000 - 0.1063 = 0.8937

5 0

24 days ago