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2 years ago

Which expressions are equivalent to 1 Over 36

Mathematics

1 answer:

Yakvenalex [24]2 years ago

7 0

Answer:

6¯²

$\frac{6^{3} }{6^{5} }$

$6^{-9}$ . $6^{7}$

its true i think

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If y=3x +2 when X = 3 find the value

Arada [10]

The correct answer is 11

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2 years ago

2(x−23)=0 plz answer

Veronika [31]

Don’t get it what r the options

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2 years ago

Jenny wants to rent struck for one day she contacted two companies Laguna truck rentals charges $20 plus $2 per mile. Salvatore

Aleks04 [339]

Ok so lets work our way up to see when the price will be the same so..luguna charges $20 plus $2 per mile and Salvatore cost 3 per mile so..

20+2=22+2=24+2=26+2=28+2=30+2=32+2=34+2=36+2=38+2=40+2=42+2=44+2=46+2=48+2=50+2=52+2=54+2

3+3=6+3=9+3=12+3=15+3=18+3=21+3=24+3=27+3=30+3=33+3+36+3=39+3=42+3=45+3+48+3=51+3=54 so it would take 18 miles hope this helped

3 0

2 years ago

You are playing a game in which a single die is rolled. If a 2 or 5 comes up, you win $36, otherwise you lose $36. What is your

ozzi

Your expected value is 36*2/6 + -36*4/6, which is equal to $-12.

7 0

2 years ago

Since the area under the normal curve within two standard deviations of the mean is 0.95, the area under the normal curve that c

alukav5142 [94]

Answer:

$X \sim N (\mu ,\sigma)$

And for this case we know this condition:

$P(\mu-2\sigma$

By the complement rule we know that:

$P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05$

But since the distribution is symmetrical we know that:

$P(X\mu +2\sigma) = 0.025$

So then the statement for this case is FALSE.

b. False

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

For this case if we define the random variable of interest X and we know that this random variable follows a normal distribution:

$X \sim N (\mu ,\sigma)$

And for this case we know this condition:

$P(\mu-2\sigma$

By the complement rule we know that:

$P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05$

But since the distribution is symmetrical we know that:

$P(X\mu +2\sigma) = 0.025$

So then the statement for this case is FALSE.

b. False

5 0

3 years ago