All categories

2 years ago

A Dodecahedron (12 sided die) is thrown, what is the P(even AND a less than or equal to 6)?

Mathematics

1 answer:

seraphim [82]2 years ago

8 0

Answer:

P = 9

Step-by-step explanation:

You might be interested in

The length of Ab & Df with work shown

Free_Kalibri [48]

It’s to blurry I can’t see

4 0

2 years ago

The equation of a graph is 4x - 3y = 5. What is the x-intercept?

user100 [1]

X intercept is where y=0
so make y=0
4x-0=5
x=5/4

4 0

2 years ago

Read 2 more answers

Write a rule that represents the function <br><br> (0,0), (1,6), (2,24), (3,54), (4,96)

Reptile [31]

These points show that this is not going to be a linear line. This is actually a parabola.

I find that the function rule is going to be

$y=6x^2$

<em>Hope this helps!!!</em>

5 0

2 years ago

Read 2 more answers

550 tickets were sold for a game for a total of $712.50. If adult tickets sold for $1.50 and children's tickets sold for $1.00,

EastWind [94]

9514 1404 393

Answer:

adult: 325
children's: 225

Step-by-step explanation:

It usually works well to let a variable represent the higher-value item in the mix. Here, we can let 'a' represent the number of adult tickets sold. Then the total revenue is ...

1.50a +1.00(550 -a) = 712.50

0.50a = 162.50 . . . . . . . . . . . . . subtract 550 and collect terms

a = 325

c = 550 -325 = 225

325 adult and 225 children's tickets were sold.

4 0

2 years ago

If log2 5 = k, determine an expression for log32 5 in terms of k.

lukranit [14]

Answer:

$log_3_2(5)=\frac{1}{5} k$

Step-by-step explanation:

Let's start by using change of base property:

$log_b(x)=\frac{log_a(x)}{log_a(b)}$

So, for $log_2(5)$

$log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)$

Now, using change of base for $log_3_2(5)$

$log_3_2(5)=\frac{log(5)}{log(32)}$

You can express $32$ as:

$2^5$

Using reduction of power property:

$log_z(x^y)=ylog_z(x)$

$log(32)=log(2^5)=5log(2)$

Therefore:

$log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)$

As you can see the only difference between (1) and (2) is the coefficient $\frac{1}{5}$ :

So:

$\frac{log(5)}{log(2)} =k\\$

$log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k$

6 0

2 years ago