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shutvik [7]
2 years ago
11

7/9 x (-5/7) Write as fraction :)

Mathematics
1 answer:
astraxan [27]2 years ago
5 0

Answer:

the answer is 35/54

Step-by-step explanation:

times the numerator and the denominator see if you can simplify it

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Which expression shows<br> another way to write (4^3)^3?
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Eight times eight equals sixty four
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The amount of money in a saving account after t years is represented by the function f(t)=3600(1.035)^t. What does the value 360
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Step-by-step explanation:

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Sarah bought a shirt on sale for $35. The original price of the shirt was three times that amount Sarah also bought a pair of sh
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Answer:

$245


Step-by-step explanation:

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2. Karin has 461 songs downloaded. Joe has
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A counselor records the number of disagreements (per session) among couples during group counseling sessions. If the number of d
Shalnov [3]

Answer:

P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)

And we can find this probability with the complement rule:

P(z>-1)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the disagreements of a population, and for this case we know the distribution for X is given by:

X \sim N(4.4,0.4)  

Where \mu=4.4 and \sigma=0.4

We are interested on this probability

P(X>4)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)

And we can find this probability with the complement rule:

P(z>-1)=1-P(z

3 0
3 years ago
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