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melisa1 [442]
2 years ago
15

Anwar selects a playing card at random from the following 6 cards: {9 of hearts, 5 of spades, 6 of hearts, 2 of spades, 4 of hea

rts, 7 of hearts} Let A be the event that Anwar selects an even-numbered card and B be the event that she chooses a heart. Which of the following statements are true?
Mathematics
1 answer:
ipn [44]2 years ago
6 0

Answer:

choose all except " The outcome of events A and B are dependent on each other"

Step-by-step explanation:

I got it right on K.A! (:

You might be interested in
Let ​ f(x)=x2+5x−36 ​. Enter the x-intercepts of the quadratic function in the boxes.
san4es73 [151]
Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

                     x^2-5*x-(36)=0 

Step by step solution:<span> Step 1:</span> Trying to factor by splitting the middle term

<span> 1.1 </span>    Factoring <span> x2-5x-36</span> 

The first term is, <span> <span>x2</span> </span> its coefficient is 1.
The middle term is, <span> -5x </span> its coefficient is  - 5.
The last term, "the constant", is <span> -36 </span>

Step-1: Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -36 = -36</span> 

Step-2: Find two factors of  -36  whose sum equals the coefficient of the middle term, which is - 5.

<span><span>     -36   +   1   =   -35</span><span>     -18   +   2   =   -16</span><span>     -12   +   3   =   -9</span><span>     -9   +   4   =   -5   That's it</span></span>


Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  4 
                     <span>x2 - 9x</span> + 4x - 36

Step-4: Add up the first 2 terms, pulling out like factors :
                    x • (x-9)
              Add up the last 2 terms, pulling out common factors :
                    4 • (x-9)
Step-5: Add up the four terms of step 4 :
                    (x+4)  •  (x-9)
             Which is the desired factorization

<span>Equation at the end of step  1  :</span> (x + 4) • (x - 9) = 0 <span>Step  2  :</span>Theory - Roots of a product :

<span> 2.1 </span>   A product of several terms equals zero.<span> 

 </span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span> 

 </span>We shall now solve each term = 0 separately<span> 

 </span>In other words, we are going to solve as many equations as there are terms in the product<span> 

 </span>Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

<span> 2.2 </span>     Solve  :    x+4 = 0<span> 

 </span>Subtract  4  from both sides of the equation :<span> 
 </span>                     x = -4 

Solving a Single Variable Equation :

<span> 2.3 </span>     Solve  :    x-9 = 0<span> 

 </span>Add  9  to both sides of the equation :<span> 
 </span>                     x = 9 

6 0
2 years ago
Read 2 more answers
Joe receives an average of 780 emails in his personal account and 760 emails in his work account each month. After changing his
Semmy [17]

Answer:

702 emails

Step-by-step explanation:

<h2>This problem bothers on depreciation of value, in this context it is Joe's email that has depreciated by 10%.</h2>

Given data

Average personal emails received monthly = 780 emails

Average work emails received monthly= 760 emails

     

      We are required to solve for the new amount of emails Joe will be receiving after changing his address, to find this value we need to solve for the depreciation of his personal mails.

      After solving for the depreciation , we then need to subtract the depreciation from the initial number of mails to get the new number of mails.

let us solve for 10% depreciation.

depreciation= \frac{10}{100} *780\\depreciation=0.1*780= 78 emails

The new number of mails

= initial number of mail- depreciation\\ =780-78= 702 emails

Joe will be receiving an average of 702 emails in his personal account monthly

7 0
2 years ago
A recent survey shows that the probability of a college student drinking alcohol is 0.6. further, given that the student is over
Murrr4er [49]

Answer:

0.24

Step-by-step explanation:

These events are not mutually exclusive; this means they can happen at the same time.

For two events A and B that are not mutually exclusive,

P(A and B) = P(A) * P(B|A)

Let A be the event "over 21 years old" and B be the event "drinks alcohol".  

The probability that a student is over 21 years old is 0.3; this is because 30% of the students are over 21 years old.

The probability that a student drinks alcohol given they are over 21 is 0.8.

This gives us

P(A and B) = 0.3(0.8) = 0.24

8 0
2 years ago
Read 2 more answers
Tickets to a show cost $5.50 for adults and $4.25 for students. A family is purchasing 2 adult tickets and 3 student tickets. If
larisa [96]

Answer:

Step-by-step explanation:

The cost of an adult ticket to the show is $5.5

The cost of a student ticket to the show is $4.25

A family is purchasing 2 adult tickets and 3 student tickets. This means that the total cost of 2 adult tickets would be

2 × 5.5 = $11

It also means that the total cost of 3 student tickets would be

3 × 4.25 = $12.75

The total cost of 2 adult tickets and 3 student tickets would be

11 + 12.75 = $23.75

If the family pays $25, the exact amount of change they should receive is

25 - 23.75 = $1.25

5 0
2 years ago
Flip two coins 100 times, and record the results of each coin toss in a table like the one below:
monitta

Answer:

1)The theoretical probability that a coin toss results in two heads showing is 25%.

2)The experimental probability that a coin toss results in two heads showing is 44%.

3) The theoretical probability that a coin toss results in two tails showing is 25%.

4) The experimental probability that a coin toss results in two tails showing is 34%.

5) The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6) The experimental probability that a coin toss results in a head and a tail is 22%.

7) The experimental probabilities are slightly different from the theoretical probabilities because the number of experiments is relatively small. As the number of experiments increase, the experimental probabilities will get closer to the theoretical probabilities.

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

Theoretical Probability:

The results you expect to happen.

Experimental Probability:

The probability determined from the result of an experiment.

1. What is the theoretical probability that a coin toss results in two heads showing?

In each toss, the theoretical  probability that a coin toss results in a head showing is 50%.

So for two coins, the probability is:

P = (0.5)^{2} = 0.25

The theoretical probability that a coin toss results in two heads showing is 25%.

2. What is the experimental probability that a coin toss results in two heads showing?

There were 100 flips, and it resulted in two heads 44 times, so:

P = \frac{44}{100} = 0.44

The experimental probability that a coin toss results in two heads showing is 44%.

3. What is the theoretical probability that a coin toss results in two tails showing?

In each toss, the theoretical  probability that a coin toss results in a tail showing is 50%.

So for two tails, the probability is:

P = (0.5)^{2} = 0.25

The theoretical probability that a coin toss results in two tails showing is 25%.

4. What is the experimental probability that a coin toss results in two tails showing?

There were 100 flips, and it resulted in two tails 34 times, so:

P = \frac{34}{100} = 0.34

The experimental probability that a coin toss results in two tails showing is 34%.

5. What is the theoretical probability that a coin toss results in one head and one tail showing?

In each toss, the theoretical probability that a coin toss results in a tail showing is 50% and in a head showing is 50%.

They can be permutated, as the tail can appear before the head, or the head before the tail. So:

P = p_{2,1}*(0.5)*(0.5) = \frac{2!}{1!}*0.25 = 0.50

The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6. What is the experimental probability that a coin toss results in one head and one tail showing?

There were 100 flips, and it resulted in a head and a tail showing 22 times, so:

P = \frac{22}{100} = 0.22

The experimental probability that a coin toss results in a head and a tail is 22%.

6 0
2 years ago
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