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White raven [17]

3 years ago

13

What is the actual distance between Saugerties and Kingston? Scale: 1 in= 2.5 mi. They are 4 in apart. What is the distance in m

iles?

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

that would be 10 miles

Step-by-step explanation:

1 = 2.5 MILES

4:x

2.5 x 4 = 10

if i am wrong please explain cause it might me being dumb or something else

You might be interested in

The times a fire department takes to arrive at the scene of an emergency are normally distributed with a mean of 6 minutes and a

Kitty [74]

Answer:

Step-by-step explanation:

Let x be the random variable representing the times a fire department takes to arrive at the scene of an emergency. Since the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 6 minutes

σ = 1 minute

the probability that fire department arrives at the scene in case of an emergency between 4 minutes and 8 minutes is expressed as

P(4 ≤ x ≤ 8)

For x = 4,

z = (4 - 6)/1 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

For x = 8

z = (8 - 6)/1 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.98

Therefore,

P(4 ≤ x ≤ 8) = 0.98 - 0.23 = 0.75

The percent of emergencies that the fire department arrive at the scene in between 4 minutes and 8 minutes is

0.75 × 100 = 75%

8 0

2 years ago

Find the distance travelled by Rubina if she takes 4 rounds of rectangular park whose length and breadth is 45m and 30m

ANTONII [103]

Length (l) = 45 m
Breadth (b) = 30 m
We know, perimeter of a rectangle = 2(length + breadth)
Therefore, perimeter of the rectangle
= 2(I + b)
= 2(45 + 30) m
= 2 × 75 m
= 150 m
So, the distance travelled by Rubina
= 4 × 150 m
= 600 m

<u>Answer:</u>

<em><u>The </u></em><em><u>distance </u></em><em><u>travelled</u></em><em><u> </u></em><em><u>by </u></em><em><u>Rubina </u></em><em><u>is </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>m.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

5 0

2 years ago

The histogram shows the weekly attendance of participants in a school's study skills program. What was the highest weekly attend

worty [1.4K]

That would be on the 5th week. There were 18 students that attended (D.)

Hope this helped!

8 0

2 years ago

Hi bro sorry to bother you but I have a question it is ok . Ok then so 875x975=

mel-nik [20]

Answer:

853,125

Step-by-step explanation:

Umhhh u can use a calculator… or write it down and slowly break down solve it like that.

Hope this helps, good luck! ;)

4 0

2 years ago

Read 2 more answers

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

2 years ago