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2 years ago

Can you evaluate (g ○ f)(0)? Explain why or why not.

Mathematics

2 answers:

chubhunter [2.5K]2 years ago

6 0

Answer:

No you can't because just like multiplication, you cant divide anything by zero.

Step-by-step explanation:

You must evaluate the function f first. Division by 0 is undefined.

Therefore, The composition cannot be evaluated.

Snowcat [4.5K]2 years ago

4 0

Answer:

sample answer on edg2020

Step-by-step explanation:

To evaluate the composition, you need to find the value of function f first. But, f(0) is 1 over 0, and division by 0 is undefined. Therefore, you cannot find the value of the composition.

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A=<3,0,-1>, find a vector b such that comp(sub a) b=2.

Anna35 [415]

Compab=a.b/|a|
b=<0,1,−2√10>
a.b= 2√10
|a| = √10
a.b/|a|=2√10 / √10=2

5 0

2 years ago

If KN =29, what is CN?

Nostrana [21]

29kn are equal to cn 2900000
Hope it was right.

6 0

2 years ago

Round off 3409725 to nearest 10

Klio2033 [76]

Answer:

10000

Step-by-step explanation:

to know if this answer is correct you have to know the rounding rules:

If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40.

If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down. Example: 33 rounded to the nearest ten is 30.

8 0

1 year ago

Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly s

kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

Case 1: P(X=1) = [P (not infected)]⁵

= (0.15 - 0.1)⁵

P(X=1) = 3.125*10⁻⁷

Case 2: P(X=6) = 1- P(X=1)

= 1 - (1 - 0.1)⁵

P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999

P(X=6) = 1.0

We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

= (1).(3.125*10⁻⁷) + 6.(1.0)

E(X) = E(X) = 6.00

Therefore, the expected number of tests, E(X) = 6.00

3 0

2 years ago

. The mean incubation time for a type of fertilized egg kept at a certain temperature is 25 days. Suppose that the incubation ti

Paul [167]

Answer:

Step-by-step explanation:

Given that the mean incubation time for a type of fertilized egg kept at a certain temperature is 25 days.

Let X be the incubation time for a type of fertilized egg kept at a certain temperature is 25 days.

X is N(25, 1)

a) Normal curve is in the attached file

b) the probability that a randomly selected fertilized egg hatches in less than 23 days

= $P(X$

we convert x into Z score and use std normal distn table to find probability

$P(X$

i.e. we can say only 2.5% proportion will hatch in less than 23 days.

8 0

1 year ago