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4 years ago

The probability distribution for a

Mathematics

1 answer:

Luden [163]4 years ago

6 0

Answer:

0.30

Step-by-step explanation:

P(x ≤ -3) = P(x=-3) + P(x=-5)

P(x ≤ -3) = 0.13 + 0.17

P(x ≤ -3) = 0.30

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What the awnser is and how to work it out because my teacher didn't show me how to

taurus [48]

The most likely answer is B.
Highlighting the corresponding parts from the original in the copy is just coloring in the parts of the copy that have the same scale, for example, if the center of the scaled copy is one unit, the original could be 3 units.

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4 years ago

number 3 says,"If u and v are the measures of complementary angles such that sin u =2/5 and tan v =(square root of 21) 21/2, lab

vodomira [7]

Answer and Explanation:

Using trig ratios, we can express the given values of sin u and tan v as shown below

$\begin{gathered} \sin u=\frac{opposite\text{ of angle u}}{\text{hypotenuse}}=\frac{2}{5} \\ \tan v=\frac{opposite\text{ of angle v}}{\text{hypotenuse}}=\sqrt[]{21} \end{gathered}$

So we can go ahead and label the sides of the triangle as shown below;

We can find the value of u as shown below;

$\begin{gathered} \sin u=\frac{2}{5} \\ u=\sin ^{-1}(\frac{2}{5}) \\ u=23.6^{\circ} \end{gathered}$

We can find v as shown below;

$\begin{gathered} u+v+90=180\text{ (sum of angles in a triangle)} \\ v+23.6+90=180 \\ v=180-113.6 \\ v=66.4^{\circ} \end{gathered}$

5 0

1 year ago

The amount of time people spend exercising in a given week follows a normal distribution with a mean of 3.8 hours per week and a

Travka [436]

Answer:

i and iii) In the figure attached part a we have the illustration for the area required for the probability of less than 2 hours and in b the illustration for the probability that X would be between 2 and 4

ii) $P(X$

And using the normal standard table or excel we got:

$P(z$

iv) $P(2$

And we can find the probability with the following difference and usint the normal standard distirbution or excel and we got:

$P(-2.25$

Step-by-step explanation:

Let X the random variable that represent amount of time people spend exercising in a given week, and for this case we know the distribution for X is given by:

$X \sim N(3.8,0.8)$

Where $\mu=3.8$ and $\sigma=0.8$

Part i and iii

In the figure attached part a we have the illustration for the area required for the probability of less than 2 hours and in b the illustration for the probability that X would be between 2 and 4

Part ii

We are interested on this probability:

$P(X$