Events [A] and [B] are independent. Find the missing probability.

PLEASE ANSWERRR!!!!!!

Fantom [35]

Answer:

C

Step-by-step explanation:

A function cannot have multiple y values for one x value.

Option A lists the y values of -8 and 10 for the x value of -4. Therefore this is not a function.

Option B lists 0, 5, 10, AND 15 all for -1. This can't be a function either.

Option C does not have two of the same x values for different results. Although both 0 and 9 give you the answer of 1 for the y value, this is perfectly okay in a function. This is your answer.

8 0

3 years ago

Can someone please tell me which one is correct??? The first screenshot is the question and the second screenshot is the options

astraxan [27]

Answer:

43.5 unit^2.

Step-by-step explanation:

Area of the triangle

= 1/2 * base * height

= 1/2 * AB * AC.

AB = √ [(6-1)^2 + (6-8)^2] = √29

AC = √ [(8 - (-7))^2 + (1 - (-5)^2] = √261

So the area = 1/2 * √29 * √261

= 43.5

7 0

1 year ago

n a group of 40 people, 10 people are healthy. The 30 unhealthy people have either high blood pressure, high cholesterol, or bot

aalyn [17]

Answer:

If a person is randomly selected from this group, the probability that they have both high blood pressure and high cholesterol is P=0.25.

Step-by-step explanation:

We can calculate the number of people from the sample that has both high blood pressure (HBP) and high cholesterol (HC) using this identity:

$N(\text{HBP or HC})=N(\text{HBP})+N(\text{HC})-N(\text{HBP and HC})\\\\\\ N(\text{HBP and HC})=N(\text{HBP})+N(\text{HC})-N(\text{HBP or HC})\\\\\\ N(\text{HBP and HC})=15+25-30=10$

We can calculate the probability that a random person has both high blood pressure and high cholesterol as:

$P(\text{HBP and HC})=\dfrac{10}{40}=0.25$

3 0

3 years ago

Need help with the rest of my problems

mariarad [96]

H) 58.48 - Instruments
I) 8.46 - Are
J) .38 - Difficult
L) 7.07 - Play
N) 22.222 - Others
O) 36.3 - Are
P) 0.18 - Cymbal

7 0

3 years ago

If an event has a 55% chance of happening in one trial, how do I determine the chances of it happening more than once in 4 trial

aev [14]

The chances of it happening more than once in 4 trials is 13%

<h3>How to determine the number</h3>

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

Learn more about probability here:

brainly.com/question/24756209

#SPJ1

6 0

2 years ago