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e-lub [12.9K]
2 years ago
11

Every day, there are 3 times more likes on an internet video of a cat which is modeled by the function c(n) = (3)n − 1, where n

is the number of days since the video posted. On the first day, there were 143 likes. What is the function that shows the number of likes each day?
A. c(n) = (3)(143)^{n-1}
B. c(n) = 143^{n-1}
C. c(n) = (3)143 − n
D.143(3)^{n-1}
Mathematics
1 answer:
Kisachek [45]2 years ago
6 0

Answer: Hi! the answer is D, now let's prove it:

Ok, let's analyze our problem; we know two facts:

1) the first day, the video has 143 likes

2) each day that pases, there are 3 times more likes.

this means, that the day 1, the video has 143 like, the day 2, has 3*143 = 429 likes, the day 3, it has 3*429 = 1287

now, this is f(n) = the number 143 multiplied by 3 (n - 1) times, where n is the amount of days.

the function that describes this is f(n) = 143*3^{n -1}

when n = 1, f(1) = 143*3^{0} = 143

when n = 2, f(2) =  143*3^{1} = 143*3

and so on, so the correct answer is D.

Also you can check the other functions if you like:

A) c(1) = (3)(143)^{1-1} = 3, so A doesn't work for the first day.

B) c(1) = 143^{1-1} = 1, B neither works for the first day.

C)  c(1) = (3)143 − 1 = 428. C neither works for the first day

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olga55 [171]

Answer: He is not correct.

Steps:

Let x1 and x2 be the first and second number, respectively.

x_2 = \frac{5}{4}x_1\\\frac{4}{5}x_2 =x_1

In words, if the second is 125%, or 5/4 of the first number (first equation),

then the first is 4/5 of the second, which is 0.8 or 80%.

8 0
1 year ago
If the odds of catching the flu among individuals who take vitamin C is 0.0342 and the odds of catching the flu among individual
melamori03 [73]

Answer:

Correct option (A).

Step-by-step explanation:

The probability of an individual catching a flu when he or she has taken vitamin C is, P (F|C) = 0.0342.

The probability of an individual catching a flu when he or she has not taken vitamin C is, P (F|C') = 0.2653.

Th ratio of individuals who caught the flu when they did not take vitamin C to those who took vitamin C is:

=\frac{P(F|C')}{P(F|C)}\\ =\frac{0.2653}{0.0342} \\=7.7573

This implies that:

P(F|C')=7.7573\times P(F|C)

Thus, the the individuals not taking vitamin C are 7.7573. Times more likely to catch the flu than individuals taking vitamin C.

The statement is True.

4 0
2 years ago
A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

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2 years ago
Eric works for an airline, and he needs to calculate the weight of passengers and luggage before takeoff. The number of
igomit [66]

Answer:

B

Step-by-step explanation:

To complete the question, here are the answer choices:

<em>A)  =A1*A2*A3*A4 </em>

<em>B)  =A1*A2+A3+A1*A4 </em>

<em>C)  =A1*(A2+A3+A1)*A4 </em>

<em>D)  =A1*A2+(A3+A1)*A4</em>

<em />

We first need to multiply A1 and A2, this will give weight of passengers.

To get weight of luggage, we multiply A1 and A4.

We also need the checked weight to add to that, which is in A3. So then we add up A3 with those 2.

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