Sam is picking fruit from a basket that contains many different kinds of fruit. Which set of events is independent?

Mathematics

1 answer:

KIM [24]3 years ago

6 0

Answer:

D.event one-He picks a kiwi and puts it back.

event two-He picks an apple and puts it back.

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Problem of the Day:

Shtirlitz [24]

Answer:

Step-by-step explanation:

What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.

in order to find this we have to split each number into it's prime components.

for 12 they are 2,2 and 3 (

⋅

)

and for 15 they are 3 and 5 (

⋅

)

Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.

Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:

girls per group, and

boys per group.

(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [

⋅

] of 3 girls [12/4] and 4 boys [16/4] )

8 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

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