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12345 [234]
3 years ago
15

Help please !!! I’m stuck on these problems

Mathematics
1 answer:
Sauron [17]3 years ago
4 0

Answer:

b=6.8\ units

Step-by-step explanation:

we know that

Applying the law of sines in the triangle ABC

\frac{AB}{sin(C)}=\frac{AC}{sin(B)}

substitute the given values

\frac{15}{sin(140\°)}=\frac{b}{sin(17\°)}

Solve for b

b=(sin(17\°))\frac{15}{sin(140\°)}

b=6.8\ units

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ahrayia [7]

Answer:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(ZStep-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 area of a population, and for this case we know the distribution for X is given by:

X \sim N(M,4)  

Where \mu=M and \sigma=4

We select a a sample of n =4 and since the distribution for X is normal then 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(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(Z

8 0
3 years ago
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