If you flip a coin ten times, how many diffrent sequences of heads and tails are possible

Whats the answer to this question?

forsale [732]

Answer:

0.67%

Step-by-step explanation:

2/300

3 0

3 years ago

Read 2 more answers

Can someone PLEASE help me with this question?

disa [49]

9514 1404 393

Answer:

11

Step-by-step explanation:

The future value of the account is given by the formula ...

A = P(1 +r/12)^(12t) . . . . principal P invested at rate r for t years

Solving for t, we find ...

A/P = (1 +r/12)^(12t) . . . . . . . . . . . divide by P

log(A/P) = 12t·log(1 +r/12) . . . . . . take logs

Divide by the coefficient of t, then fill in the numbers.

t = log(A/P)/(12·log(1 +r/12)) = log(202800/93000)/(12·log(1 +.068/12))

t ≈ 11.497

It will take about 11 years for the account balance to reach the desired amount.

4 0

3 years ago

(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct

Luba_88 [7]

Splitting up the interval of integration into $n$ subintervals gives the partition

$\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]$

Each subinterval has length $\dfrac{1-0}n=\dfrac1n$ . The right endpoints of each subinterval follow the sequence

$r_i=\dfrac in$

with $i=1,2,3,\ldots,n$ . Then the left-endpoint Riemann sum that approximates the definite integral is

$\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n$

and taking the limit as $n\to\infty$ gives the area exactly. We have

$\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}$

6 0

3 years ago

A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o

Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(X $E(X_{k}) = n_{k} * p_{k}$

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = $\frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}$

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= $\frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\$ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= $\frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\$ = 0.000001

c.

P(X₁≤2) = $\frac{22!}{0!} * 0.08^{0} * (0.92)^{22}$ + $\frac{22!}{1!} * 0.08^{1} * (0.92)^{21}$ + $\frac{22!}{2!} * 0.08^{2} * (0.92)^{20}$ = 0.7442

I hope you have a SUPER day!

8 0

3 years ago

kayden paddled a canoe to an island. the island is 8 miles from the shore. his trip to the island took two hours while paddling

Andrew [12]

Answer:

$2\text{ mph}$

Step-by-step explanation:

GIVEN: kayden paddled a canoe to an island. the island is $8$ miles from the shore. his trip to the island took two hours while paddling against the current. he paddles $6$ mph with no current.