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Dennis_Churaev [7]
3 years ago
11

5^11+5^10 over 5^10-5^8?

Mathematics
2 answers:
Kitty [74]3 years ago
7 0

Answer:

6.25

Step-by-step explanation:

5^11 + 5^10 =  58593750

5^10 - 5^8 = 9375000

58593750 / 9375000

= 6.25


guajiro [1.7K]3 years ago
6 0

5^11 + 5^10

-------------------     =

5^10 - 5^8


48,828,125 + 5^10

--------------------------    =

5^10 - 5^8


48,828,125 + 9,765,625

---------------------------------     =

5^10 - 5^8


48,828,125 + 9,765,625

-----------------------------------    =

9,765,625 - 5^8


48,828,125 + 9,765,625

---------------------------------    =

9,765,625 - 390,625


58,593,750

------------------------------   =

9,765,625 - 390,625

58,593,750

------------------  =

9,375,000

answer: 6.25

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Answer:

x=0 or x= -3/2

Step-by-step explanation:

Step 1: Subtract 11x from both sides.

24x4+92x−11x=11x−11x

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Step 2: Factor left side of equation.

3x(2x+3)(4x2−6x+9)=0

Step 3: Set factors equal to 0.

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Assume the upper arm length of males over 20 years old in the United States is approximately Normal with mean 39.3 centimeters (
oksian1 [2.3K]

Answer:

a) For this case we can use this:

P(X

P(X>\mu +3*\sigma)=P(X>46.5)=0.0015

So the range of lengths would be 32.1 and 46.5 since on the middle of these two values we have 1-0.0015-0.0015=0.997 or 99.7 % of the values.

b) P(X>36.9)

And using the complement rule we have:

P(X>36.9)= 1-P(X

Step-by-step explanation:

Previous concepts

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the variable of interest.

From the problem we have the mean and the standard deviation for the random variable X. E(X)=39.3, Sd(X)=2.4

So we can assume \mu=39.3 , \sigma=2.4

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

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• The probability of obtain values within three deviation's from the mean is 0.997

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P(X    

P(X>\mu +\sigma)=P(X >41.7)=0.16  

P(X    

P(X>\mu +2*\sigma)P(X>44.1)=0.025

P(X

P(X>\mu +3*\sigma)=P(X>46.5)=0.0015

Part a

For this case we can use this:

P(X

P(X>\mu +3*\sigma)=P(X>46.5)=0.0015

So the range of lengths would be 32.1 and 46.5 since on the middle of these two values we have 1-0.0015-0.0015=0.997 or 99.7 % of the values.

Part b

For this casee we want this probability:

P(X>36.9)

And using the complement rule we have:

P(X>36.9)= 1-P(X

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inna [77]

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8 0
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(You plug in the ordered pairs in the inequality and then check if the statement is true)
8 0
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