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3 years ago

Fellow person with an easy question here :))

Mathematics

1 answer:

Leviafan [203]3 years ago

3 0

The actual wingspan of fly is 5.5cm

Step-by-step explanation:

Given,

Scale of drawing;

4mm = 1cm

Ratio of mm to cm = 4:1

Wingspan of fly in drawing = 22mm

Let,

x be the actual wingspan.

Ratio of mm to cm = 22:x

Using proportion;

Ratio of mm to cm :: Ratio of mm to cm

$4:1::22:x\\$

Product of extreme = Product of mean

$4*x=22*1\\4x=22$

Dividing both sides by 4

$\frac{4x}{4}=\frac{22}{4}\\x=5.5$

The actual wingspan of fly is 5.5cm

Keywords: proportion, division

Learn more about division at:

#LearnwithBrainly

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Pls somebody solve this I still don’t understand it yet :/ please

Fofino [41]

Answer:

first blank 39, second 9/39 i think, and third 351

Step-by-step explanation:

i haven't done something like this in a long time so i dont know if its completely correct or correct at all

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2 years ago

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago

Suppose that lithium calculator battery with a charge of 3 volts, actually has a voltage with uniform distribution between 2.93

Dmitrij [34]

Answer:

b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

$M = \frac{a + b}{2}$

The standard deviation of the uniform distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

Uniform distribution between 2.93 and 3.1 volts

This means that $a = 2.93, b = 3.1$ . So

Mean:

$M = \frac{2.93 + 3.1}{2} = 3.015$

Standard deviation:

$S = \sqrt{\frac{(3.1 - 2.93)^{2}}{12}} = 0.049$

What is the probability that a battery has a voltage less than 2.98?

$P(X \leq 2.98) = \frac{2.98 - 2.93}{3.1 - 2.93} = 0.2941$

So the correct answer is:

b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

6 0

3 years ago

Does anyone know this??

HACTEHA [7]

Answer: 0.34

Explanation: First, you add all of the probabilities together to find the total probability when Dr. Green is missing. (You will get 0.66.) Next, you will need to subtract this from 1.00. 1 - 0.66 = 0.34. 0.34 is your final answer.

8 0

2 years ago

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If a cone shaped vase can hold 45 cubic centimeter of water and has a radius of 2 centimeters what is the height of the vase use

xz_007 [3.2K]

Answer:

The height of the vase is $h=10.75\ cm$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

we have

$V=45\ cm^{3}$

$r=2\ cm$

$\pi =3.14$

substitute and solve for h

$45=\frac{1}{3}(3.14)(2)^{2} h$

$135=(3.14)(4)h$

$135=12.56h$

$h=135/12.56$

$h=10.75\ cm$

3 0

3 years ago