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2 years ago

Square root of 16r raised to the 6

1 answer:

7 0

If you mean: (√(16r)^6 that is the same as:

(16r)^(1/2)^6

relevant rule: (a^b)^c=a^(b*c), in this case

(16r)^(6*1/2)

(16r)^3

4096r^3

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The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0

3 years ago

Great Amusements Park has been raising its ticket prices every year, as shown in the table below

Fiesta28 [93]

1.)

Between year 0 and year 1, we went from $50 to $55.

$55/$50 = 1.1

The price increased by 10% from year 0 to year 1.

Between year 2 and year 1, we went from $55 to $60.50.

$60.50/$55 = 1.1

The price also increased by 10% from year 1 to year 2. If we investigate this for each year, we will see that the price increases consistently by 10% every year.

The sequence can be written as an = 50·(1.1)ⁿ

2.) To determine the price in year 6, we can use the sequence formula we established already.

a6 = 50·(1.1)⁶ = $88.58

The price of the tickets in year 6 will be $88.58.

4 0

3 years ago

Let θ

konstantin123 [22]

Answer:

θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Step-by-step explanation:

Given :

Length of side opposite to angle θ is y

Length of side adjacent to angle θ is x

θ is part of a right angle triangle

At this instant,

x = 8 , $\frac{dx}{dt}$ = 7

( $\frac{dx}{dt}$ denotes the rate of change of x with respect to time)

y = 8 , $\frac{dy}{dt}$ = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = $\frac{y}{x}$ -------------------------------------------------------------------1

At this instant,

tanθ = $\frac{8}{8}$ = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or $\frac{d}{dt}$ (θ)

$\frac{d}{dt}$ (tanθ) = $\frac{d}{dt}$ (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

$sec^{2}$ θ $\frac{d}{dt}$ (θ) = $\frac{1}{x}$ $\frac{dy}{dt}$ - $\frac{y}{x*x}$ $\frac{dx}{dt}$ -----------------------2

Substituting the values of x , y , $\frac{dx}{dt}$ , $\frac{dy}{dt}$ , θ at that instant in equation (2)

2 $\frac{d}{dt}$ (θ) = $\frac{1}{8}$ *(-14)- $\frac{8}{8*8}$ *7

$\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Therefore θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

3 0

3 years ago

What is the result when the number 16 is decreased by 25%? Show me your work plz

adoni [48]

Answer:

Step-by-step explanation:

25% of 16 is : 16 ÷ 100 × 25 = 4

16 decreased by 4 is = 12

8 0

3 years ago

State the amplitude and period of f(x) = 3cos9/5t

Leno4ka [110]

Answer:

The amplitude is 3 and the period is 10π/9.

Step-by-step explanation:

The standard cosine function is in the form:

$y=a\cos(b(x-c))+d$

Where |a| is the amplitude, 2π/b is the period, c is the phase shift, and d is the vertical shift.

We have the function:

$f(x)=\displaystyle 3\cos\frac{9}{5}t$

We can rewrite this as:

$\displaystyle f(x)=(3)\cos\left(\frac{9}{5}\left(t-0\right)\right)-(0)$

Therefore, a = 3, b = 9/5, c = 0, and d = 0.

Hence, our amplitude is |3| = 3.

Our period will be:

$\displaystyle \text{Period}=\frac{2\pi}{9/5}=2\pi \cdot \frac{5}{9}=\frac{10\pi}{9}$

7 0

3 years ago