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vlada-n [284]
3 years ago
6

Determine whether each of these functions from {a,b,c,d} to itself is one-to-one. a) f(a) = b, f(b) = a, f(c) = c, f(d) = d. b)

f(a) = b, f(b) = b, f(c)=d, f(d) = c. c) f(a) = d, f(b) = b, f(c)=c, f(d) = d.
Mathematics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

The first one is one-to-one.

The second one is not one-to-one.

Third one is not one-to-one.

The problem:

Are the following one-to-one from {a,b,c,d} to {a,b,c,d}:

a)

f(a)=b

f(b)=a

f(c)=c

f(d)=d

b)

f(a)=b

f(b)=b

f(c)=d

f(d)=c

c)

f(a)=d

f(b)=b

f(c)=c

f(d)=d

Step-by-step explanation:

One-to-one means that a y cannot be hit more than once, but all the y's from the range must be hit.

So the first one is one-to-one because:

f(a)=b

f(b)=a

f(c)=c

f(d)=d

All the elements that got hit are in {a,b,c,d} and all of them were hit.

The second one is not one-to-one.

The reason is because f(a) and f(b) both are b.

Third one is not one-to-one.

The reason is because f(a) and f(d) are both d.

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An ice chest contains cans of apple juice, cans of grape juice, cans of orange juice, and cans of mango juice. Suppose that you
mel-nik [20]

Answer:

4.9%

Step-by-step explanation:

This question is incomplete. However; it can be found on search engines. The complete question is  as follows :

An ice chest contains cans of six apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting three cans of grape juice.

Solution :

In an ice chest there are different cans of juice. Among them

Number of cans of apple juice = 6

Number of cans of grape juice = 8

Number of cans of orange juice = 4

Number of cans of mango juice = 2

Total number of cans of juice = 6 + 8 + 4 + 2 = 20

Let A, B and C are the event of selecting of three cans.  The events A, B and C are dependent.

Probability of selecting three cans of juice

P = \frac{\text{number of grape cans}}{\text {total number of cans}}

P (A) = \frac{8}{20}

P (B) = \frac{7}{19}

P (B) = \frac{6}{18}

P = \frac{8}{20} × \frac{7}{19} × \frac{6}{18}

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Probability of selecting three cans of grape juice is 4.9%

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Find the maximum value of y = -2(x - 4)2 - 6. A) -6 B) -4 C) -2 D) 4
otez555 [7]

Y=-2(x-4)^2-6

The -6 makes it shift from origin(0,0) to (0,-6)

Then the -4 inside with x, makes it shift to the right by for, so the center would be now instead of (0,0), (4,-6).

As it has a negative A factor, it is a parabola open downwards, so the center is the maximum value.

Answer: (4,-6)

Hope you get it!

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