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swat32
3 years ago
5

Write six-thousandths in standard form

Mathematics
2 answers:
fenix001 [56]3 years ago
8 0

Answer:

6x10^{3}

Step-by-step explanation:

We are asked to write six thousandths in standard form.

First of all, we will write our given number as a decimal. Our given number is six thousands, so we will divide 6 by 1000 as:

6x10^{3}

To write 0.006 as standard form, we need two factors whose product is 0.006. First factor would be any number between 1 to 10 and the second number would be power of 10.

Since we have 3 decimals before 6, so our first factor would be 6 and second factor would be .

Therefore, our required number would be .

Brums [2.3K]3 years ago
7 0

Answer:

.006

Step-by-step explanation:

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Answer:

1. 1244.07

2. 653.45

3. 706.86

4. 141.37

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6. 402.12

7. 863.94

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Step-by-step explanation:

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2 years ago
The power P produced by a wind turbine is directly proportional to the cube of the wind speed S. A wind speed of 37 miles per ho
vfiekz [6]

Answer:the output for a wind speed of 40 miles per hour is 810.8 kilowatts.

Step-by-step explanation:

The power P produced by a wind turbine is directly proportional to the cube of the wind speed S. This means that

P is directly proportional to S

Introducing a constant of proportionality, k, it becomes

P = kS

A wind speed of 37 miles per hour produces a power output of 750 kilowatts. This means that

750 = k × 37

k = 750/37 = 20.27

The expression becomes

P = 20.27S

Therefore, the output for a wind speed of 40 miles per hour would be

P = 20.27 × 40

P = 810.8

6 0
3 years ago
Part A.) A garden hose supplies 36 gallons of water in 3 minutes. Use a table of equivalent ratios to show the garden house's wa
saw5 [17]

Answer: It would be 360 gallons of water per 10 mins because

Step-by-step explanation:

You multiply 36×10 and you get 360 gallons

7 0
3 years ago
Write an equation in point-slope form for the line that has a slope of 5 and contains the point (-1,2)
Cloud [144]

Answer:

y - 2 = 5(x+1)

Step-by-step explanation:

Point slope form is

y-y1 = m(x-x1)  where m is the slope and ( x1,y1) is a point on the line

y - 2 = 5(x- -1)

y - 2 = 5(x+1)

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2 years ago
Read 2 more answers
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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