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2 years ago

The variance Var(x) for the binomial distribution is given by equation :

Mathematics

1 answer:

aliina [53]2 years ago

6 0

Answer: d. np(1 - p).

Step-by-step explanation:

Let x be any binomial variable which represents the number of success such that $X\sim B(n, p)$ , where n is the sample size or the total number of trials and p is the probability of getting success in each trial .

Then, the mean E(x) and the variance Var(x) for the binomial distribution is given by equation :

$E(x)=\mu=np$

$Var (x)=\sigma^2=np(1-p)$

where n is the sample size or the total number of trials and p is the probability of getting success in each trial .

Therefore , the correct option is option d. np(1 - p) .

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Peter wants to fence the park in front of his home on three sides which measure 152m 40cm, 205m 10cm, 310m 39cm. What is the tot

Tpy6a [65]

One centimenter is 0.01 meters. So, you can write the measures as

$152.40\text{m},\quad 205.10\text{m},\quad 310.39\text{m}$

Once this rewriting is done, getting the total length $T$ is quite trivial, since all measurements are in the same unit, and we can simply sum everything:

$T = 152.40+205.10+310.39 = 667.89$

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2 years ago

A quality-control inspector is testing a batch of printed circuit boards to see wheater they are capable of performing in a high

avanturin [10]

Answer:

see explaination

Step-by-step explanation:

Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.

a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.

b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.

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2 years ago

A survey of over 7,000 randomly selected employees in 15 nations recently showed that employees who work

arlik [135]

Answer:

Not necessarily, because this wasn't an experiment.

Step-by-step explanation:

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2 years ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

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2 years ago

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

Dovator [93]

Option a: $\frac{m^{5} }{162n}$ is the equivalent expression.

Explanation:

The expression is $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ where $m\neq 0, n\neq 0$

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

$\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }$

Cancelling the like terms, we have,

$\frac{3^{-3}m^{5} n^{-1}}{6 }$

This equation can also be written as,

$\frac{m^{5}}{3^{3}6 n^{1} }$

Multiplying the terms in denominator, we have,

$\frac{m^{5} }{162n}$

Thus, the expression which is equivalent to $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ is $\frac{m^{5} }{162n}$

Hence, Option a is the correct answer.

8 0

2 years ago

Read 2 more answers