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4 years ago

Find (f*f)(3) f(x)=x^2-x A. 30 B. 6 C. 33 D. -6

Mathematics

1 answer:

Tems11 [23]4 years ago

6 0

Answer:

Step-by-step explanation:

$(f\circ f)(x)=f\bigg(f(x)\bigg)\\\\f\bigg(f(3)\bigg)\\\\\text{Calculate}\ f(3):\\\\f(3)=3^2-3=9-3=6\\\\f\bigg(f(3)\bigg)=f(6)=6^2-6=36-6=30$

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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 parallelogram is formed by the supports that attach a basketball backboard and rim to the wall. The angles change as the baske

irina [24]

The measure of angle BCD is 32 degrees

<h3>How to determine the value of BCD?</h3>

The image of the parallelogram that completes the question is added as an attachment.

The given parameter is:

m<DAB = 32 degrees

From the attached image, we can see that BCD and DAB are opposite angles.

Opposite angles of a parallelogram are equal.

So, we have:

BCD = 32 degrees

Hence, the measure of BCD is 32 degrees

Read more about parallelograms at:

brainly.com/question/21322889

#SPJ1

3 0

2 years ago

X f(x)<br> -2 -1/200<br> -1 -1/20<br> 0 -1/2<br> 1 -5<br> 2 -50

Hitman42 [59]

9514 1404 393

Answer:

f(x) = -1/2·10^x

Step-by-step explanation:

The y-intercept is -1/2.

The common ratio is (-1/20)/(-1/200) = 10, so the exponential function can be written as ...

f(x) = -1/2·10^x

6 0

3 years ago

Plz answer my question it is urgent..!!

oee [108]

Answer:

In the step-by-step explanation!

Step-by-step explanation:

Not sure if it is too late but here:

1.)

$\frac{1}{\alpha}+\frac{1}{\beta} \\\frac{\beta}{\alpha \beta } +\frac{\alpha}{\alpha \beta } \\\frac{\alpha +\beta }{\alpha \beta }$

2.)

$\frac{1 }{\alpha } *\frac{1}{\beta} = \frac{1*1}{\alpha*\beta} =\frac{1}{\alpha \beta}$

3.)

$\frac{1}{\alpha}-\frac{1}{\beta}\\ \frac{\beta}{\alpha \beta } -\frac{\alpha }{\alpha \beta } \\\frac{\beta -\alpha }{\alpha \beta }$

Hopes this help! Please give me Brainliest!

6 0

3 years ago

4. Is rap music more popular among young Indians than among young Asians? A sample survey compared 634 randomly chosen Indians a

bulgar [2K]

Answer:

H0 is accepted

there is no difference between the proportions of Indian and Asian young people who listen to rap music every day.

Step-by-step explanation:

Given a sample survey compared 634 randomly chosen Indians aged 15 to 25 with 567 randomly selected Asians in the same age group.

$p_{1} = \frac{368}{634} =0.580$

It found that 368 of the Indians and 130 Asians listened to rap music every day.

$p_{2} = \frac{130}{567} =0.229$

Null hypothesis H0: there is no difference between the proportions of Indian and Asian young people who listen to rap every day.

$p_{1} = p_{2}$

Alternative hypothesis:- $p_{1} \neq p_{2}$

Level of significance α = 0.05

The test of statistic

$z = \frac{p_{1} - p_{2}}{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} }) } } }$

where $p = \frac{n_{1}p_{1} + n_{1}p_{2}}{n_{1}+n_{2}} = \frac{634(0.580)+567(0.229)}{634+567}$

p = 0.414

and q = 1-p = 1- 0.414 =0.586

$Z = \frac{0.580-0.229}{\sqrt{0.414(0.586)(\frac{1}{634} +\frac{1}{567} } } }$

on calculation , we get

z = 0.300 ><1.96 at 95 % level of significance

H0 is accepted

there is no difference between the proportions of Indian and Asian young people who listen to rap every day.

3 0

3 years ago