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monitta
3 years ago
9

What does the denominator of the fraction 2/6 mean.

Mathematics
2 answers:
ryzh [129]3 years ago
5 0

Answer:

The denominator of a fraction would be the lower part of the fraction. If you were to simplify this fraction it would be 1/3  

Step-by-step explanation:

salantis [7]3 years ago
4 0
The whole like parts of whatever the thing is?
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I think it would be D but i am not for sure

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Is this a function help
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3 years ago
SBC is an isosceles triangle with main vertex S.
stira [4]

Answer + Step-by-step explanation:

1) D be the symmetric of B with respect to C then CD = BC

A the symmetric of C with respect to B then AB = BC

We obtain :

CD = BC

AB = BC

Then AB = CD

2) m∠SBA = 180 - SBC = 180 - SCB = m∠SCD

3) we have :

BA = CD

BS = CS

m∠SBA = m∠SCD

Then

the triangles SBA and SCD are congruent

4)

the triangles SBA and SCD are congruent Then SA = SD

Therefore SAD is an isosceles triangle.

4 0
2 years ago
Which of the following is a solution of y &gt; Ixl - 6?<br> O (5, -1)<br> O (-1,-5)<br> O (-5, 1)
Ghella [55]

Answer:

i think its the first option

Step-by-step explanation:

8 0
3 years ago
The distribution of the number of people in line at a grocery store has a mean of 3 and a variance of 9. A sample of the numbers
Airida [17]

Answer:

a) P(\bar X >4)=P(Z>\frac{4-3}{\frac{3}{\sqrt{50}}}=2.357)

P(Z>2.357)=1-P(Z

b) P(\bar X \frac{2.5-3}{\frac{3}{\sqrt{50}}}=-1.179)

P(Z

c)  

P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}}

P(-1.179Step-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".  

Part a

Let X the random variable that represent the number of people of a population, and for this case we know that:

Where \mu=3 and \sigma=\sqrt{9}=3

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P(\bar X >4)= P(z> \frac{4-3}{\frac{3}{\sqrt{50}}})

And using a calculator, excel or the normal standard table we have that:

P(Z>2.357)=1-P(Z

Part b

P(\bar X  \frac{2.5-3}{\frac{3}{\sqrt{50}}})

And using a calculator, excel or the normal standard table we have that:

P(Z

Part c

For this case we want this probability:

P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}}

And using a calculator, excel or the normal standard table we have that:

P(-1.179

7 0
3 years ago
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