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Alex787 [66]
3 years ago
14

Answer all please.....^

Mathematics
1 answer:
postnew [5]3 years ago
8 0
#4) 1/6
#5) 1/2
#6) 4/15
#7) 8/15
#8) 11/15
#9) 12
#10) 15
#11) 112

Explanation
#4) There is one section marked 5 out of 6 sections.

#5) There are three odd-numbered sections out of 6 sections.

#6) 2 blue were tossed 4 times out of 15 tosses.

#7) 1 blue and 1 pink were tossed 8 ties out of 15 tosses.

#8) All blue was tossed 4 times out of 15; this means that all blue was not tossed 15-4=11 times out of 15.

#9) There are 4 choices for location and 3 choices for transportation; 4(3) = 12.

#10) There are 3 choices for level and 5 choices for character; 3(5) = 15.

#11) Since 30% are rock, 100%-30% = 70% are not rock.  70% = 70/100 = 0.7; 0.7*160 = 112.
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X + 2y = -3 x - y = -12
Nastasia [14]

Answer:

x = -9, y =3

Step-by-step explanation:

x + 2y = -3

x - y = -12

Subtract to two equations to eliminate x

x + 2y = -3

-x + y = 12

---------------------

   3y = 9

Divide each side by 3

3y/3 = 9/3

y = 3

Now we can find x

x-y =-12

x - 3 = -12

Add 3 to each side

x-3+3 = -12+3

x = -9

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The product of two rational numbers is -14/45. If one of the no. is -7/9 find the other no.
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The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th
Nikitich [7]

Answer:

a) \mu_{\bar X} =13.35

e.none of the above

b) \sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02

a.0.0200

c) z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5

P(\bar X< 13.32)= P(z

a.0.0668

d) z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5

z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5

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c.0.6853

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".  

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".

Solution to the problem

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

X \sim N(13.35,0.12)  

Where \mu=13.35 and \sigma=0.12

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}})

Part a

\mu_{\bar X} =13.35

Part b

\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02

Part c

We want this probability:

P(\bar X< 13.32)

For this case we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got:

z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5

P(\bar X< 13.32)= P(z

Part d

We want this probability:

P(13.30

For this case we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got:

z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5

z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5

P(13.30

8 0
3 years ago
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