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3 years ago

4. (03.05)

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

5 0

Answer:

B. domain is all integers 0 through 9

Step-by-step explanation:

Domain values area usually plotted along the horizontal axis. In this case, the number of days is the domain of the of the function.

The domain of the function that is plotted on the graph ranges from 0, 1, 2 to 9.

1, 2, 3 to 9 are all integers. Therefore, we can conclude that the "domain is all integers 0 through 9."

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F (x) = 2x^2 - 4x + 1

Kobotan [32]

Answer:

<h2><u>tinanong ko rin po yang tanong nyo at ito po ang sagot nila</u></h2>

3 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. A group of 4 independent samples are checked.

galben [10]

Answer:

a) 0.6561

b) 0.2916

c) 0.3439

Step-by-step explanation:

We are given the following information:

Let us treat high level of contamination as our success.

p = P(High level of contamination) = P(success) = 0.10

n = 4

The, by binomial distribution:

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}\\\text{where x is the number of success}$

a) P(No high level of contamination)

We put x = 0, in the formula.

$P(X=0) = \binom{4}{0}.(0.10)^0.(1-0.10)^{4} =0.6561$

Probability that no lab specimen contain high level of contamination is 0.6561

b) P(Exactly one high level of contamination)

We put x = 1, in the formula.

$P(X=1) = \binom{4}{1}.(0.10)^1.(1-0.10)^{3} =0.2916$

Probability that no lab specimen contain high level of contamination is 0.6561

c) P(At least one contains high level of contamination)

$p(x \geq 1) = 1 - p( x = 0) = 1 - 0.6561 = 0.3439$

Probability that at least 1 lab specimen contain high level of contamination is 0.3439

8 0

3 years ago

URGENT!! Trigonometry practice question!!

marusya05 [52]

Answer:

what did u get the answers

5 0

3 years ago

1. What is the simple interest on $4,500 for 2 and a half years at 4% per year?

allochka39001 [22]

Answer: 1st is 180$ sorry if I’m wrong

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Answer:

emmainna [20.7K]

Let the numbers be n, n+2, n+4

Sum equals too= 13+2(n+4), which is 2n+21

a) Equation--> n+n+2+n+4= 2n+21

b) Solution--> 3n+6= 2n+21

=> n= 15

c) Second number--> 17 (15+2)

Third number--> 19 (15+4)

d) 15+15+2+15+4=30+21

=> 51= 51

So, the equation is true.

And pls mark me brainliesttt :)))

8 0

2 years ago