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3 years ago

Give the formula for a function with domain (5, ∞) that has the following properties:

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

7 0

Answer:

The area under the curve y = f(x) at x-axis, on the interval x E (5, ∞) is equal to 7.

Step-by-step explanation:

Solution

Given that:

f(x) = 7√x-5

The function of domain is:

D = x-5 > 0

x>5

so,

x← (5, ∞)

Now,

(1) The sequence root function is always known as positive, i.p defined from x E (5, ∞)

i.p =√x-5

Thus,

7/√x-5 >0

Therefore f(x) = 7/√√x-5 which is a positive integer, which is defined in domain x E (5, ∞).

Each value exist a real and unique values of f(x)

Now,

The function f(x) is continuous and over the interval(5, ∞)

Note: Kindly find an attached copy of part to the solution of this given question below

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Randall recorded 8 songs on his most recent CD the total length of the CD is 49 minutes find a until rate to represent the avera

olganol [36]

Answer:

6.125 minutes per song

Step-by-step explanation:

Unit rates are based on a single unit of something. Like price per 1 gallon of gas; price for 1 pound of bananas; miles driven per 1 gallon of gas, etc. What we are looking for in minutes per song.

Use a proportion to solve this with minutes on top and number of songs on the bottom. We know that 49 minutes = 8 songs, and we want to know how many minutes (x) per 1 song:

$\frac{min}{song}:\frac{49}{8}= \frac{x}{1}$

Cross multiply to get

8x = 49 so

x = 6.125 minutes per song

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3 years ago

5000 jumps for 7 hours how many per hour

xenn [34]

Here is the formula.

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1 year ago

Approximately 40% of the calls to an airline reservation phone line result in a reservation being made. (round to three decimal

IRISSAK [1]

Answer:

The probability that none of the 10 calls result in a reservation is 0.60%. In turn, the probability that at least one call results in a reservation being made is 99.40%.

Step-by-step explanation:

Since approximately 40% of the calls to an airline reservation phone line result in a reservation being made, supposing an operator handles 10 calls, to determine what is the probability that none of the 10 calls result in a reservation, and what is the probability that at least one call results in a reservation being made, the following calculations must be performed:

0.6 ^ 10 = X

0.006 = X

0.006 x 100 = 0.60%

Therefore, the probability that none of the 10 calls result in a reservation is 0.60%.

100 - 0.60 = 99.40

In turn, the probability that at least one call results in a reservation being made is 99.40%.

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3 years ago

(3 points) Two friends went fishing. Hilda caught a 16 inch rainbow trout and Bran caught a 23 inch brown trout. Suppose rainbow

wolverine [178]

We are assuming here that the lengths of rainbow trout and brown trout follow a normal distribution, that is, both lengths are normally distributed (which is reasonable for this kind of random variable in populations, like the heights in humans, for example).

Answer:

The rainbow trout resulted longer relative to the corresponding type of fish, that is, one standard deviation above the mean or longer than the 84.134% of the population for rainbow trouts because the brown trout resulted in 0.6 standard deviations above the population for brown trouts (or 72.575% above them).

Step-by-step explanation:

Because both types of fish come from different species, they cannot be fairly compared using the normal distribution of each fish.

Instead, we can compare them if we "transform" these raw data into z-scores in a way that we can use standardized values to determine how far from the mean are the lengths for each fish.

In other words, we have to remember that z-scores tell us the distance of a raw value from the population mean in standard deviations units, no matter the parameters of the normal distribution they coming from. A positive value for the z-score tells us that the raw score is above the population mean, whereas a negative value tells us that the raw score is below it.

We have to remember that the normal distribution is determined by two parameters, namely, the population mean and the population standard deviation.

In the question, we have two lengths coming from two different normal distributions. We can transform them into standardized values so that we can use the standard normal distribution, and then obtain how far from the mean are the lengths for each fish and the percentile that corresponds to this z-score.

The formula for the z-score is

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

x is the raw score. In this case, we have two. One is the length of the rainbow trout (16 inches) and the brown trout length (23 inches).

$\\ \mu$ is the population mean for each normal distribution. We also have two. For the rainbow trout mean length, we have a mean of 12 inches, and a mean length of 20 inches for the brown trout.

$\\ \sigma$ is the population standard deviation for each normal distribution. A standard deviation for the rainbow trout of 4 inches, and a standard deviation of 5 inches for the brown trout.

Having all this information at hand, we can proceed to solve the question.

Z-score for rainbow trout

For this case, we have

x = 16 inches.

$\\ \mu = 12$ inches.

$\\ \sigma = 4$ inches.

Using formula [1]

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{16 - 12}{4}$

$\\ z = \frac{4}{4}$

$\\ z = 1$

That is, the length of the rainbow trout caught by Hilda is one standard deviation above the mean. If we consult a cumulative standard normal table, we have that

$\\ P(z$

In words, the length for the fish that Hilda caught is, approximately, above than 84.134% of the population for rainbow trouts (one standard deviation above the population mean).

Z-score for brown trout

We can proceed in the same way as we did with the previous case.

For this case, we have

x = 23 inches.

$\\ \mu = 20$ inches.

$\\ \sigma = 5$ inches.

Using formula [1]

$\\ z = \frac{23 - 20}{5}$

$\\ z = \frac{3}{5}$

$\\ z = 0.6$

That is, the length of the brown trout caught by Bran is 0.6 standard deviations units above the mean. With this value at hand, we can conclude that this fish is smaller relative to the type of fish (brown trout) than the rainbow trout caught by Hilda. We can confirm this using the standard normal table:

$\\ P(z$

In words, the length for the fish that Bran caught is, approximately, above than 72.575% of the population for brown trouts (or the length of 72.575% of the population for brown trout are below the length of the brown trout caught by Bran).

In conclusion, the rainbow trout resulted longer relative to the corresponding type of fish, that is, one standard deviation above the mean or longer than the 84.134% of the population for rainbow trouts, because the brown trout resulted in 0.6 standard deviations above the population for brown trouts (or 72.575% above them).

In the graph below, a standard normal distribution (mean = 0, standard deviation = 1), we can confirm the previous result (the red color for the rainbow trout, and the blue color for the brown trout).

8 0

3 years ago