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sp2606 [1]
3 years ago
5

4.78 times 10 to the 6th power was the largest attendance record of the polar bear express. chilly football extravaganza. 3.98 t

imes 10 to the 6th power attended at the last last pb express event. how many more people were in attendance on the record-setting day?
Mathematics
1 answer:
Gennadij [26K]3 years ago
6 0

9514 1404 393

Answer:

  8×10⁵ = 800,000

Step-by-step explanation:

The difference is ...

  (4.78×10⁶) -(3.98×10⁶) = 0.80×10⁶ = 8×10⁵

8×10⁵ = 800,000 more people attended at the largest extravaganza.

_____

Your calculator (or a spreadsheet) can do the subtraction and display the number in whatever format you want.

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Nastasia [14]

Answer:

the measure of the exterior angle is 87 degrees.

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2 years ago
3/4 2/3 5/6 5/7 least to greatest
Lana71 [14]

Answer:

2/3, 5/7, 3/4, 5/6.

Step-by-step explanation:

Convert each one to a decimal fraction:

3/4 = 0.75

2/3 = 0.666..

5/6 = 0.8333..

5/7 = 0.714

So least to greatest is

2/3, 5/7, 3/4, 5/6.

6 0
3 years ago
Read 2 more answers
Worth 12 points and please actually help me
Alekssandra [29.7K]

Answer:

(-3,0)

Step-by-step explanation:

Step 1 identify coordinates of A

The coordinate of A is (-6 , 2)

Step 2 apply translation by adding 3 to the x value and subtracting 2 from the y value

(-6 + 3 , 2 - 2)

Simplify

(-3 , 0)

6 0
3 years ago
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
you are allowed to work a total of no more than 30 hours each week at your two jobs. Lawn mowing pays $5 per hour and babysittin
timama [110]

Answer:

5X + 8Y >= 300;    intersection at  (-20, 50)

Step-by-step explanation:

let t = work hours

0 < t < 30

X = time lawn mowing

Y = time babysitting.

X + Y < 30

5X + 8Y >= 300

We could solve...

X < 30 - Y

5(30 - Y) + 8Y  >=300

150 - 5Y + 8Y >= 300

3Y >=150

Y >=50

then  X < -20  

intersection at  (-20, 50)

4 0
3 years ago
Read 2 more answers
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