MATHEMATICS COLLEGE

What is the area of this triangle?

Answers

Answer 1

Answer:

Answer:

Option (D)

Step-by-step explanation:

Formula for the area of a triangle is,

Area of a triangle = $(1)/(2)(\text{Base})(\text{Height})$

For the given triangle ABC,

Area of ΔABC = $(1)/(2)(\text{AB})(\text{CD})$

Length of AB = $(y_2-y_1)$

Length of CD = $(x_3-x_1)$

Now area of the triangle ABC = $(1)/(2)(y_2-y_1)(x_3-x_1)$

Therefore, Option (D) will be the answer.

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A roll of ribbon was 9 meters long. Kevin cut 8 pieces of ribbon each of length 0.8 meter, to tie some presents. He then cut the remaining ribbon into some pieces, each of length 0.4 meter.a) how many pieces of ribbon, each 0.4 meter in length, did Kevin have? b) What was the length of ribbon left over?
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The value of the Euler \phi function (\phi is the Greek letter phi) at the positive integer n is defined to be the number of positive integers less than or equal to n that are relativel prime to n, For example for n=14, we have {1, 3, 5, 9, 11, 13} are the positive integers less than or equal to 14 which are relatively prime to 14. Thus \phi (14) = 6.Find the following:\phi(9) = __________.\phi(15) = __________.\phi(75) =__________.

Answers

Answer:

$\bf \phi (9)=5$

$\bf \phi (15)=8$

$\bf \phi (75)=40$

Step-by-step explanation:

Positive integers relative primes to 9 which are less or equal than 9

{1,2,4,5,7}

$\bf \phi (9)=5$

Positive integers relative primes to 15 which are less or equal than 15

{1,2,4,7,8,11,13,14}

and

$\bf \phi (15)=8$

Positive integers relative primes to 75 which are less or equal than 75

{1,2,4,7,8,11,13,14,16,17,19,22,23,26,28,29,31,32,34,37,38,41,43,44,46,47,49,52,53,56,58,59,61,62,64,67,68,71,73,74}

and

$\bf \phi (75)=40$

As the saying goes, “You can't please everyone.” Studies have shown that in a largepopulation approximately 4.5% of the population will be displeased, regardless of the
situation. If a random sample of 25 people are selected from such a population, what is the
probability that at least two will be displeased?
A) 0.045
B) 0.311
C) 0.373
D) 0.627
E) 0.689

Answers

The probability that at least two people will be displeased in a random sample of 25 people is approximately 0.202.

What is probability?

It is the chance of an event to occur from a total number of outcomes.

The formula for probability is given as:

Probability = Number of required events / Total number of outcomes.

Example:

The probability of getting a head in tossing a coin.

P(H) = 1/2

We have,

This problem can be solved using the binomialdistribution since we have a fixed number of trials (selecting 25 people) and each trial has two possible outcomes (displeased or not displeased).

Let p be the probability of an individual being displeased, which is given as 0.045 (or 4.5% as a decimal).

Then, the probability of an individual not being displeased is:

1 - p = 0.955.

Let X be the number of displeasedpeople in a random sample of 25.

We want to find the probability that at least two people are displeased, which can be expressed as:

P(X ≥ 2) = 1 - P(X < 2)

To calculate P(X < 2), we can use the binomial distribution formula:

$P(X = k) = (^n C_k) * p^k * (1 - p)^(n-k)$

where n is the samplesize (25), k is the number of displeasedpeople, and (n choose k) is the binomial coefficient which represents the number of ways to choose k items from a set of n items.

For k = 0, we have:

$P(X = 0) = (^(25)C_ 0) * 0.045^0 * 0.955^(25)$

≈ 0.378

For k = 1, we have:

$P(X = 1) = (^(25)C_1) * 0.045^1 * 0.955^(24)$

≈ 0.42

Therefore,

P(X < 2) = P(X = 0) + P(X = 1) ≈ 0.798.

Finally, we can calculate,

P(X ≥ 2) = 1 - P(X < 2)

= 1 - 0.798

= 0.202.

Thus,

The probability that at least two people will be displeased in a random sample of 25 people is approximately 0.202.

Learn more about probability here:

brainly.com/question/14099682

#SPJ2

Answer:

Step-by-step explanation:

The correct answer is (B).

Let X = the number of people that are displeased in a random sample of 25 people selected from a population of which 4.5% will be displeased regardless of the situation. Then X is a binomial random variable with n = 25 and p = 0.045.

P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – binomcdf(n: 25, p: 0.045, x-value: 1) = 0.311.

P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – 0C25(0.045)0(1 – 0.045)25 – 25C1(0.045)1(1 – 0.045)24 = 0.311.

My soccer team has 16 players. I have to choose a starting lineup of a goalie and 10 regular players (the regular players are interchangeable). How many different starting lineups can i choose? *the answer is not 4368*

Answers

Answer:

3003

Step-by-step explanation:

We want to find out how many ways we can choose 10 players among 15 players (since the goalie is not interchangeable)

The number of different lineups you can have can be found by using combination:

$^(15)C_(10) = (15!)/((15 - 10)! 10!)\n \n= (15!)/(5! * 10!) \n\n= 3003$

There are 3003 different lineups that can be chosen.

Final answer:

To determine the number of starting lineups, we use combinations in probability. We first choose a goalie from 16 players, then 10 regular players from the remaining 15, giving us 48048 unique lineups.

Explanation:

This problem can be solved by using the concept of combinations in probability and statistics. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and ! denotes factorial, which is the product of all positive integers up to that number.

Firstly, we need to choose a goalie. There are 16 players, so the number of ways to choose a goalie is C(16, 1) = 16.

After choosing the goalie, we are left with 15 players. Then we need to choose 10 players to fill in the rest of the team. Thus, the number of ways to choose the 10 regular players is C(15, 10).

The total number of unique starting lineups is then the product of these two results. Hence, the solution would be C(16, 1) * C(15, 10) = 16 * 3003 = 48048 different starting lineups.

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300 people joined the community orchestra. Some intended to play a string instrument, some a brass instrument, and the others percussion. If 3 times as many people were planning to play a string instrument as a brass instrument, and if 60 fewer were going to play percussion than a brass instrument, how many people were planning to play each type of instrument?

Answers

Total amount of instrument players= 300
Number of brass instrument players= x (because we do not know how many there are, except that all other values that we must calculate are based off of the number of brass players we have
Number of string instruments= 3x (because 3 times more than the number of brass instruments
Number of Percussion= x-60 (because there are 60 less percussion players than brass players

So, our equation is (number of string instruments)+(number of brass instruments)+(number of percussion instruments)=300 (which is our total possible number of people

3x+x+x-60=300
3x+x+x=360
5x=360
x= 72

We have 72 brass instrument players, 3(72)= 216 string instrument players, and 72-60=12 percussion players

check: 72+216+12=300

Hey can you please help me posted picture of question

Answers

Vertical stretching a function means multiplying it by a constant.

So, vertical stretching of F(x) by a factor of 7 will be equal to 7 F(x).

Thus,

G(x) = 7 F(x)

G(x) = 7x²

So, option C gives the correct answer

One normally distributed with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between 34.5 and 35 in.

Answers

Answer:

The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Step-by-step explanation:

Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z represents the overlap.

It is provided that:

$X\sim N(20,\ 0.50^(2))\nY\sim N(15,\ 0.40^(2))\nZ\sim N(1,\ 0.10^(2))$

It is provided that the lengths and amount of overlap are independent of each other.

Compute the mean and standard deviation of total length as follows:

$E(T)=E(X+Y-Z)\n=E(X)+E(Y)-E(Z)\n=20+15-1\n=34$

$SD(T)=√(V(X+Y-Z))\n=√(V(X)+V(Y)+V(Z))\n=\sqrt{0.50^(2)+0.40^(2)+0.10^(2)}\n=0.6480741\n\approx 0.65$

Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.

$T\sim N(34, 0.65^(2))$

Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:

$P(34.5<T<35)=P((34.5-34)/(0.65)<(T-\mu_(T))/(\sigma_(T))<(35-34)/(0.65))\n\n=P(0.77<Z<1.54)\n\n=P(Z<1.54)-P(Z<0.77)\n\n=0.93822-0.77935\n\n=0.15887\n\n\approx 0.1589$

*Use a z-table.

Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

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