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4 years ago

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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.

1 answer:

vfiekz [6]4 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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3 years ago

The two boxes shown are cuboids.

kramer

Answer:

250 small cubes

Step-by-step explanation:

Because the volume of the bigger cube 30000

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3 years ago

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 2643 miles, with a standard

musickatia [10]

Answer:

$P(2643-51< \bar X < 2643+51)= P(2592< \bar X$

And we can use the z scoe formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limits we got:

$z = \frac{2592-2643}{\frac{368}{\sqrt{44}}}= -0.919$

$z = \frac{2694-2643}{\frac{368}{\sqrt{44}}}= 0.919$

And this probability is equivalent to:

$P(-0.919$

Step-by-step explanation:

For this case we can define the random variable X as "number of miles between services" and we know the following info given:

$\mu = 2643 , \sigma = 368$

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We select a random sample size of n =44. And we want to find this probability:

$P(2643-51< \bar X < 2643+51)= P(2592< \bar X$

And we can use the z scoe formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limits we got:

$z = \frac{2592-2643}{\frac{368}{\sqrt{44}}}= -0.919$

$z = \frac{2694-2643}{\frac{368}{\sqrt{44}}}= 0.919$

And this probability is equivalent to:

$P(-0.919$

4 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that $\mu = 50$ and $\sigma = 6$ .

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

$Z = \frac{X - \mu}{\sigma}$

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0

3 years ago

A school ends the year with 651 students. This is a 5% increase from the start of the school year.

Jet001 [13]

ANSWER:

n = 1.05n divided by 105 times 100

OR

n = (the number of students at the END of the school year) divided by 105 times 100

WORKING OUT:

1.05n = 651 students
1n = 651/105 times 100 = ANSWER

5 0

3 years ago

Read 2 more answers