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2 years ago

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PLEASE HELP DUE TODAY ILL GIVE BRAINLIEST 1. An advertising company charges $40 per half-page advertisement and $75 per full-pag

e advertisement. Michael has a budget of $800 to purchase 13 advertisements. (a) Define a variable for each unknown. Write a system of equations to represent the situation. (b) How many half-page advertisements does Michael purchase? Show your work. (c) How many full-page advertisements does Michael purchase? Show your work.

1 answer:

Oksana_A [137]2 years ago

8 0

Answer: inverse of given function = log(x)/log(3)

Step-by-step explanation: how to find the inverse of any function?

in the complete function, replace y with "x" and x with "y"then, solve the obtained equation from step 2 for yand, then the value obtained for "y" is the inverse required i.e. = f^{-1}(x)

now, we solve the problem:

given : - y = 3^x

replacing the value of "y" with "x", we get,

x = 3^y

to solve for "y" we should take the "logarithm" for both sides

log (x) = y log(3)

or y = log(x)/log(3)

or f^(-1) (x) = log(x)/log(3)

( I hope this helped <3 )

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Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec

jeka94

Answer:

a) Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

Defect >2 = cannot be sold

Y = the number of defects on a fat quarter that can be sold by Company F.

Y = defect that can be sold

Y = Defect less or equal to 2 = 0,1,2

Probability distribution of the random variable Y:

Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean of Y (μ)

μ = Σ x*P(Y)

= (0*0.58) +(1*0.23)+(2*0.11)

= 0+0.23+0.22 = 0.45

Standard deviation of Y = σ

σ = Σ√(x-mean)^2*P(Y)

= Σ√[(x- μ )^2*P(Y)]

= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]

= √[0.11745 + 0.069575 +0.264275

= √(0.4513

σ = 0.6718

Company G:

σ for defect that be sold = 0.66

μ for defect that be sold = 0.40

Difference between μ of F and μ of G

= 0.45-0.40 = 0.05

Difference between σ of F and σ of G

= 0.67-0.66 = 0.01

Selling price of fat quarter without defect = $5

Discount per defect = $1.5

Selling price per defect = 5-1.5 = $3.5

Discount per 2 defect = $1.5*2 = $3

Selling price per defect = 5-3 = $2

Since defect to be sold cannot be greater than 2, let Y = 5,3,2

Probability distribution of the selling price Y:

Y 5 3 2

P(Y) 0.58 0.23 0.11

μ = (5*0.58) +(3.5*0.23)+(2*0.11)

μ = 2.9+0.805+0.22 =1.285

σ = Σ√[(x- μ )^2*P(Y)]

σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]

σ = 8.00+0.68+0.06 = 8.74

7 0

2 years ago

Mrs.Steﬀen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin

qaws [65]

Answer:

a. $\\ 10! = 3628800$ ;

b. $\\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}$

Step-by-step explanation:

We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.

The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

Thus, we need to multiply the possibilities after each selection: that is <em>why</em> the <em>Multiplication Principle</em> is important here.

This could be expressed mathematically using n!:

$\\ n! = n * (n-1)! * (n-2)! *...* 2*1.$

For instance, $\\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.$

So, for the case in question, the <em>ten</em> students can line up in:

$\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800$ ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy <em>only</em> the following places in forming the file:

$\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th}$ <em>places</em>.

The members of Group 2 <em>only</em>:

$\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th}$ <em>places</em>.

And the members of Group 3, the following <em>only</em> ones:

$\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th}$ <em>places.</em>

Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.

After using the <em>Multiplication Principle</em>, we have, thus:

$\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19}$ <em>ways the students can line up to come in from recess</em>.

3 0

1 year ago

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equ

kompoz [17]

Answer:

0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution.

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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 800, \sigma = 40, n = 16, s = \frac{40}{\sqrt{16}} = 10$

Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

This probability is the pvalue of Z when $X = 775$ . So

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

$Z = \frac{775 - 800}{10}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

5 0

2 years ago

A bag contains one red pen, four black pens, and three blue pens. two pens are randomly chosen from the bag and are not replaced

bagirrra123 [75]

8 total pens....4 are black

first pick, probability of being black is 4/8
2nd pick. without replacing, probability is 3/7

probability of a black pen picked first and then another black pen picked again is : 4/8 * 3/7 = 12/56 = 0.21

8 0

1 year ago

Read 2 more answers

On a coordinate plane, a cube root function goes through (negative 2.5, negative 3.5), crosses the y-axis at (0, negative 3), ha

mart [117]

Answer:

$g(x) = \sqrt[3]{x-1} - 2$

Step-by-step explanation:

We want to find h and k in:

$g(x) = \sqrt[3]{x-h} + k$

At the inflection point, the second derivative is equal to zero, so:

$g'(x) = \frac{1}{3} (x-h)^{\frac{-2}{3}}$

$g''(x) = \frac{1}{3} \frac{-2}{3}(x-h)^{\frac{-5}{3}} = 0$

Then x - h = 0.

Inflection point is located at (1, -2), replacing this x value we get:

1 - h = 0

h = 1

We know that the point (-2.5, -3.5) belongs to the function, so:

$-3.5 = \sqrt[3]{-2.5-1} + k$

k ≈ -2

All data, used or not, are shown in the picture attached.

4 0

2 years ago