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3 years ago

8

We know that logarithmic functions are a bit more unusual than functions we’ve seen in the past, such as linear, exponential, qu

adratic, and even polynomial and radical functions. We rarely see them in examples, so what’s their real purpose as functions?

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

Step-by-step explanation:

can you re-phrase that please then i can answer

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The drama club was selling ticketsto the school play. Adult ticketscost $8.00 each, and studenttickets cost $5.00 each. The litt

BlackZzzverrR [31]

We are given a problem that can be solved using a system of linear equations. Let A, be the number of adults, and S the number of students. Since there are in total 142 people and there were two days, this means that the sum of the number of adults and the number of students must be 284, which can be written mathematically as follows:

$A+S=284,(1)$

This is our first equation. The second equation is found using the total sales of $1948. Since the ticket per adult is $8 and per student is $5, we have the following equations:

$8A+5S=1948,(2)$

To solve this equation we will solve for A in equation (1), by subtracting S to both sides;

$\begin{gathered} A+S-S=284-S \\ A=284-S \end{gathered}$

Now we will replace this value in equation (2):

$8(284-S)+5S=1948$

Now we will apply the distributive property:

$2272-8S+5S=1948$

Addins like terms:

$2272-3S=1948$

Subtracting 2272 to both sides;

$\begin{gathered} 2272-2272-3S=1948-2272 \\ -3S=-324 \end{gathered}$

Dividing both sides by -3:

$S=-\frac{324}{-3}=108$

Now we replace this value in equation (1), where we have already solved for A:

$\begin{gathered} A=248-108 \\ A=140 \end{gathered}$

Therefore, there were sold 108 student tickets and 140 adult tickets.

3 0

1 year ago

Rob and Amy each improved their yards by planting rose bushes and shrubs. They bought their supplies from the same store. Rob sp

Dmitriy789 [7]

Answer:

The cost of one rose bushes be $10 and the cost of shrubs $4.

Step-by-step explanation

Let us assume that the cost of rose bushes be x .

Let us assume that the cost of shrubs be y .

As given

Rob and Amy each improved their yards by planting rose bushes and shrubs.

They bought their supplies from the same store.

Rob spent $100 on 8 rose bushes and 5 shrubs.

Than the equation

8x + 5y = 100

Amy spent $112 on 8 rose bushes and 8 shrubs.

8x + 8y = 112

Than the two equation

8x + 5y = 100 and 8x + 8y = 112

Subtracting 8x + 5y = 100 from 8x + 8y = 112

8x - 8x + 8y - 5y = 112 - 100

3y = 12

$y = \frac{12}{3}$

y = $4

Put in the equation 8x + 5y = 100 .

8x + 5 × 4 = 100

8x + 20 = 100

8x = 100 - 20

8x = 80

$x = \frac{80}{8}$

x = $10

Therefore the cost of one rose bushes be $10 and the cost of shrubs $4.

7 0

4 years ago

Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ma

SOVA2 [1]

Answer:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=10$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size

Solution to the problem

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The sample mean $\bar X$ is distributed on this way:

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

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$

4 0

3 years ago

A phone company charges a base fee of $25.00 per

iogann1982 [59]

Answer:

c-25/ m

hope that helps

Step-by-step explanation:

Algebreic manipulation, watch Khan acadmy.

8 0

3 years ago

The ratio of men to women working for a company is 3 to 2 . If there are 80 employees total, how many men work for the company?

djyliett [7]

sorry i messed up its actually 48 men

7 0

3 years ago