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swat32
2 years ago
7

Can we combine different functions (quadratic, linear, exponential, cube root) into one piecewise function? If the answer is yes

, then why?
Plzzzz help
Mathematics
1 answer:
yanalaym [24]2 years ago
8 0

Answer: yes, because it is useful to describe changes in the behavior of the thing we are describing.

Step-by-step explanation:

A piecewise function f(x) is a function that is "different" depending on the range of the values of x, this is for example:

y = x -----> if x < 0

y = x^2 ----- if x ≥ 0

This is used to represent changes in the behavior of the thing we are describing.

For example, suppose an object that is moving with constant velocity, the velocity graph will be a constant line.

Now, the object suddenly, at a time t = t0 accelerates to a constant acceleration A, now the graph of the velocity is a linear graph with slope equal to A.

So this situation can be represent this with a piecewise function:

V(t) = c -----> if t < t0

V(t) = c + A*t -----> if t ≥ t0

So the answer is yes, we can.

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Consider 8 blood donors chosen randomly from a population. The probability that the donor has type A blood is .40. Which of the
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Answer: Option B is the only correct option.

Step-by-step explanation:

Number of samples = n = 8.

Probability of success = p = 0.4

Probability of failure = q = 0.6

r = chosen number of donors among the 8

To solve this question, we use the distribution formula

P(x=r) = nCr * p^r * q^n-r

For option A, to check if P(3<x<5) = 0.37. [3 and 5 inclusive]

When x = 3

P(x=3) = 8C3 * 0.4^3 * 0.6^5

P(x=3) = 56 * 0.064 * 0.07776

P(x=3) = 0.2787

When x= 4

P(x=4) = 8C4 * 0.4^4 * 0.6^4

P(x=4) = 70 * 0.0256 * 0.1296

P(x=4) = 0.2322

Since p(x=3) + p(x=4) is already greater than 0.37, then we know option A is NOT correct.

For option B, To check if the probability of 1 or fewer donor is about 0.11. i.e if P(x</=1) = 0.11

When x=o

P(x=0) = 8C0 * 0.4^0 * 0.6^8

P(x=0) = 1* 1 * 0.016796

P(x=0) = 0.016796.

When x = 1

P(x=1) = 8C1 * 0.4^1 * 0.6^7

P(x=1) = 8 * 0.4 * 0.02799

P(x=1) = 0.08958

P(x=0) + P(x=1) = 0.016796 + 0.08958

P(x=0) + P(x=1) = 0.10635.

Since this is approximately 0.11, then option B is a correct option.

For option C to check if the probability 7 or more donors not having type A = 0.0087

To do this,we determine thw probability of 7 or more donors having type A and we subtract our answer from 1.

First, we determine P(x>/=7)

When x= 7

P(x=7) = 8C7 * 0.4^7 * 0.6^1

P(x=7) = 8 * 0.001638 * 0.6

P(x=7) = 0.007864

When x=8

P(x=8) = 8C8 * 0.4^8 * 0.6^0

P(x=8) = 1 * 0.0006554 * 1

P(x=8) = 0.0006554

P(x=7) + P(x=8) = 0.007864 + 0.0006554 = 0.00852.

Since probability of 7 or more donors having type A is 0.00852 as against what was stated in the option C, then option C is NOT a correct option.

For option D, to check if the probability of exactly 5donors having type A blood = 0.28

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P(x=5) = 8C5 * 0.4^5 * 0.6^3

P(x=5) = 56 * 0.01024 * 0.216

P(x=5) = 0.1239.

Since probability of what was derived for having exactly 5 donors having sample A is different from what wqs given in the option, then option D is NOT correct.

For option E, since what was stated in the option negates what was derived for exactly 5 donors, then option E is NOT correct

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