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Masja [62]
3 years ago
7

BRAINLIEST GOES TO THE CORRECT ANSWER

Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0
I wish i knew sorry
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Suppose that Y takes on the values {0, 1, 2, 3}, each with probability 1/4. Take a random sample of 5 numbers from this distribu
arsen [322]

Answer:

The total number of samples that give this outcome is 5.

Step-by-step explanation:

Since Y takes values in {0,1,2,3}, For us to have that Y_1+Y_2+Y_3+Y_4+Y_5=1 implies that all of them are zero but one. The one that is non-zero necessarily is equal to 1. To calculate the number of samples that give this outcome is equivalent to counting the total number of ways in which we can pick the i-index such that Y_i=1. Note that in this case we can either choose Y1 to be 1, Y2 to be 1 and so on. So, the total number of samples that give this outcome is 5.

8 0
3 years ago
The value of a piece of property is growing at a rate of 3.5% annually if the property was worth $77,000 when it was purchased,
lilavasa [31]
The answer is .C (1.035)x
4 0
3 years ago
Is my answer correct? If its not, would you mine telling me what is correct?
Luden [163]

Answer:

Sorry but you have the y and x mixed up

Step-by-step explanation:

You have them mixed up

6 0
3 years ago
Read 2 more answers
Name each polynomial<br><br> 4m^2-5m+6<br><br> 18s-7<br><br> 6p^3
faltersainse [42]

Answer:

Quadratic function

linear function

cubic function

Step-by-step explanation:

5 0
3 years ago
A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03​% chlorine. Starting at tequals​0, city water conta
Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is  

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.  

Remember, 1 h = 60 minutes. The initial value problem is  

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x =  - 6* 10⁻⁴*(3x - 10⁻¹)               x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us  

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C    or    |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).  

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then  

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If  t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0
3 years ago
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