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mars1129 [50]
2 years ago
8

Kyle notices that the number of likes he receives each day on his social media page is a function. He decides to create a graph

representing the feedback over one month. So far he has collected the following data: (0, 0), (1, 12), and (2, 24), where x represents days and y represents likes.
How many likes would Kyle receive on the 30th day?
A) 30
B) 96
C) 144
D) 360
Mathematics
1 answer:
nadya68 [22]2 years ago
5 0

Answer:

D) 360

Step-by-step explanation:

The data Kyle collected is (0, 0), (1, 12), and (2, 24), where x represents days and y represents likes.

We can write a linear function to model this situation.

This will be of the form

y = mx

because it passes through the origin , where m is the slope.

The slope is

m =  \frac{12 - 0}{1 - 0}  = 12

Hence the equation is

y = 12x

To find the number of likes on the 30th day, we put x=30 into the equation to get:

y = 12 \times 30 = 360

The correct choice is D

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Urgent help needed...............
andreyandreev [35.5K]

Answer:

Option b is correct

\{x | x \neq \pm 7, x\neq 0\}.

Step-by-step explanation:

Domain is the set of all possible values of x where function is defined.

Given the function:

h(x) = \frac{9x}{x(x^2-49)}

To find the domain of the given function:

Exclude the values of x, for which function is not defined

Set denominator = 0

x(x^2-49) = 0

By zero product property;

x = 0 and x^2-49= 0

⇒x = 0 and  x^2 =49

⇒x = 0 and x = \pm 7

Therefore, the domain of the given function is:

\{x | x \neq \pm 7, x\neq 0\}

7 0
3 years ago
Read 2 more answers
Plz help me well mark brainliest if you are correct!
evablogger [386]
The answer is D. ..........
4 0
2 years ago
15 points and Will give branliest !!
prisoha [69]

Answer:

Midpoint of a line segment with the endpoints A(a1, a2), B(b1, b2) is M(1/2*(a1 + b1), 1/2*(a2 + b2))

So,

(1/2*(-4 + 7), 1/2*(-3 + (-5))) = (1/2*3, 1/2*(-8)) = (1.5, -4)

Answer D is correct.


8 0
3 years ago
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
2 years ago
Mr. P mowed 2/7 of his lawn. His son mowed 1/4 of it who mowed the most how much of the lawn still need to be mowed. Solve using
Agata [3.3K]

Answer:

Step-by-step explanation:

1/3

7 0
2 years ago
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