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ser-zykov [4K]
3 years ago
9

What is the product written in scientific notation? 6.5x10^-7 4x10^-4

Mathematics
1 answer:
Pavel [41]3 years ago
8 0
Remember that scientifc notation is always (number greater than or equal to 1 and less than 10) times 10^(number of decimal places moved to the right)

ok so remember htat whe multiplying you can assosiat the numbesr so we have (6.5)(10^-7)(4)(10^-4)=(6.5 times 4)(10^-7 times 10^-4)
also remember the exponential law that says
x^m times x^n=x^(m+n) so

10^-7 times 10^-4=10^(-7+-4)=10^-11

so now we have
6.5 times 4 times 10^-11
26 times 10^-11
26 needs to be more than 1 and less than 10
2.6 times 10^1 times 10^-11
combine the 10's
10^1 times 10^-11=10^(1+-11)=10^-10
2.6 times 10^-10
answer is 2.6 times 10^-10
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What is the slope of the line that passes through the points (2,1) and (-4,-5)
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\text{Slope =} \dfrac{1+5}{2+4} =   \dfrac{6}{6} = 1
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Sal the sloth takes one hour to travel 7 1/2 feet. If he continues to travel at the same rate, how long will it take him to trav
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2 years ago
(3 points) Blades of grass Suppose that the heights of blades of grass are Normally distributed and independent, with each heigh
NemiM [27]

The final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

What is the normal distribution?

A continuous probability distribution for a real-valued random variable in statistics is known as a normal distribution or Gaussian distribution.

If x follows a normal distribution with mean μ and standard deviation σ then the distribution of

\sum_{i =1}^{n}x_{i}  follows an approximately normal distribution with a mean n\mu and standard deviation \sqrt{n }\sigma

let x be the height of blades of grass

x follows normal distribution with mean = μ = 4 and standard deviation = σ = 0.75.

Y = x1 + x2 +...........+x10

Y = \sum_{i =1}^{10}x_{i}

Distribution of Y is normal with,

Mean = \mu _{y}=10*4 = 40 and standard deviation = \sigma _{y}=\sqrt{10}*0.75 = 2.3717

a)

P( Y < 42.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

=NORMDIST( x, mean, SD , 1 )      

=NORMDIST(42.5, 40, 2.3717, 1 )

=0.8541

P( Y < 42.5 )  = 0.8541

b)

P( 39.5 < Y < 40.5 ) = P( Y < 40.5 ) - P( Y < 39.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

P( Y < 40.5 )  =NORMDIST(40.5, 40, 2.3717, 1 ) = 0.5835

P( Y < 39.5 ) = NORMDIST(39.5, 40, 2.3717, 1 ) = 0.4165

P( 39.5 < Y < 40.5 ) = 0.5835 - 0.4165  = 0.1670

P( 39.5 < Y < 40.5 ) = 0.1670

Hence, the final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

To learn more about the normal distribution visit,

brainly.com/question/4079902

#SPJ4

5 0
1 year ago
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