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4 years ago

PRE-CAL HELP

Mathematics

1 answer:

BabaBlast [244]4 years ago

7 0

Be
P = (21, 12)
Q = (19, -8).
The vector u will be given by
u = PQ = Q-P = (19, -8) - (21, 12) = (- 2, -20)
The magnitude of this vector is
u = ROOT ((2 ^ 2) + (20 ^ 2)) = 20.1
Therefore, the vector v as it is in the opposite sense to u will be given by
v = 0.4i + 4j.
Its magnitude is
v = Root((0.4^2)+(4^2))=4.02

Dividing both magnitudes it is verified that the magnitude of u is five times greater than that of v
20.1 / 4.02 = 5.
Finally, the correct answer is
v = 0.4i + 4j

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4 years ago

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a) $P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

That correspond to approximate 6.44%

b) $P(X>31)=P(\frac{X-\mu}{\sigma}>\frac{31-\mu}{\sigma})=P(Z>\frac{31-26.3}{5.4})=P(Z>0.87)$

And we can find this probability using the normal standard table or excel and we got:

$P(Z>0.87)=1-P(Z$

That correspond to approximate 19.2%

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".

Part a

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

$X \sim N(26.3,5.4)$

Where $\mu=26.3$ and $\sigma=5.4$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

That correspond to approximate 6.44%

Part b

We are interested on this probability

$P(X>31)$

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:

$P(X>31)=P(\frac{X-\mu}{\sigma}>\frac{31-\mu}{\sigma})=P(Z>\frac{31-26.3}{5.4})=P(Z>0.87)$

And we can find this probability using the normal standard table or excel and we got:

$P(Z>0.87)=1-P(Z$

That correspond to approximate 19.2%

7 0

3 years ago