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atroni [7]
3 years ago
12

Kimi’s dog gives birth to a litter of 4 puppies per year, on average. The relationship between time and the expected total numbe

r of puppies is shown on the graph. Which other ordered pairs would fall on this line? Select all that apply.
(0, 0)
(2, 6)
(3, 12)
(4, 14)
Mathematics
2 answers:
Daniel [21]3 years ago
4 0

Answer:

answer is (0, 0) & (3, 12)

Step-by-step explanation:

taking the text right now

denis-greek [22]3 years ago
3 0

Answer:a,c

Step-by-step explanation:

You might be interested in
A = 18, b = 12, A = 27°
Daniel [21]

Answer:

Could you post an image?

I have the variables; I just need the problem.

3 0
3 years ago
5) Mr. Morales drove his motorcycle 258
xxMikexx [17]

Answer:

43 m/hr

Step-by-step explanation:

258m/6hr = 43m/hr

distance over time

6 0
3 years ago
The median of $20, x, 15, 30, 25$ is $0.4$ less than the mean. If $x$ is a whole number, what is the sum of all possible values
34kurt

Answer:

198

Step-by-step explanation:

5 0
3 years ago
Which quadratic function has a leading coefficient of 2 and a constant term of –3?
Romashka [77]
The answer is <span>f(x) = 2x2 + 3x – 3
</span>
f(x) = ax² + bx + c
a - the leading coefficient
c - the constant term

<u>We are looking for a = 2, c = -3</u>

Through the process of elimination:
The first (f(x) = 2x3 – 3) and the third choice (f(x) = –3x3 + 2) have x³ so these are not quadratic function.

In the function: <span>f(x) = –3x2 – 3x + 2
</span>a = -3
c = 2

In the function: f(x) = 2x2 + 3x – 3
a = 2
c = -3
6 0
3 years ago
Read 2 more answers
4. At a local university 54.3% of incoming first-year students have computers. If 3 students are selected at random, and the fol
valentinak56 [21]

Answer:

a) 0.0954

b) 0.9045

c) 0.1601

Step-by-step explanation:

We are given the following information:

We treat first-year students having computers as a success.

P(first-year students have computers) = 54.3% = 0.543

Then the number of first year students follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

a) P(None have computers)

P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954

b) P(At least one has a computer)

P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045

c) P(All have computers)

P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601

3 0
3 years ago
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