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True [87]
3 years ago
14

Which of the following events has a probability of 75% or higher?

Mathematics
1 answer:
Cloud [144]3 years ago
7 0

Answer:

D

Step-by-step explanation:

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Help find the answer for problem number 4
Olin [163]
Adult ticket (a) = $5
Child ticket (c) = $2

785 tickets = $3280

a + c = 785 tickets
5a + 2c = $3280

c = 215 child tickets
a = 570 adult tickets

570 + 215 = 785 tickets
5(570) + 2(215) = $3280

There were 215 child tickets sold on Saturday



3 0
3 years ago
Suppose you have $23,000 that you are going to invest at a 7% interest rate compounded quarterly. According to the compound inte
andrew11 [14]
YAAS! I literally learned this just a few weeks ago. (I am not actually a high school senior) First, convert 7% to a decimal, which is 0.07. Then multiply 0.07 by 23,000, giving you 1,610. This is your unit rate. I will though, need you to be more specific on "compounding quarterly" because I actually do not know what that means.
5 0
3 years ago
Sandra is a real estate agent and rams 5000 commission on the selling of a 62,000 house. what is the percent of commission she r
sveta [45]
To find a percent, you use this simple formula:
5000/62000 = x/100

In this case, x is going to be your percent of commission.

So, to find this, you first divide 5000 by 62000, which is about equal to 0.08.

Now your equation looks like this:
0.08 = x/100

Next, you multiply both sides by 100 to reverse the operation. Now, your equation will look like this:

8 = x

So, the answer is 8%.
8 0
3 years ago
Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
3 years ago
Given cos 0 = 11/6 and angle 0 is in Quadrant I, what is the exact value of sin 0 in
user100 [1]

1 = sin²0 + cos²0

sin²0 = 1 - cos²0

sin²0 = 1 - 11/36

sin²0 = 25/36

sin 0 = 5/6 or -5/6

In the first quadrant, the values for sin, cos and tan are positive.

sin0 = 5/6

8 0
3 years ago
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