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Kryger [21]
3 years ago
9

Gary’s is a small publishing company that publishes math books. The production costs include a one-time cost for editing the boo

k, which is $1,050 for the book the company is currently working on (The Universe of Math). Once the book is in production, the only cost will be the variable cost of printing, which is $15 per book. The company will market the book at a price of $35. What equation represents the profit earned by selling x books?
Mathematics
1 answer:
sveta [45]3 years ago
5 0
Y=20x -1500, y equals the prophet, x is the number of books, 20 is the money made after taking out the cost of production, 1500 is a cost that only occurs once.
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Find the volume of the prism. <br> 4.8 m<br> 10 m<br> 7.2 m
kondaur [170]

Answer:

The volume of the prism is 22m

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3 years ago
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HELP MATH QUESTIONS!! WILL MARK CORRECT ANSWERS BRAINLIEST!! 40 POINTS!! DUE TODAY ASAP!!
yuradex [85]

Answer:

  1. f(x) = 34.59 +3.74x
  2. 0.96
  3. very strong
  4. 8.93

Step-by-step explanation:

1. The selection of answers offers a choice of positive or negative slope, a slope of about 4 or about 35, and y-intercepts from the same set of values.

If you consider a couple of points (at the ends of the domain is usually a good choice), you see the slope is about 4 and the y-intercept is around 30.

For example, the points (5, 50) and (15, 91) have a slope of (91-50)/(15-5) = 4.1 so will give rise to the equation ...

  y -50 = 4.1(x -5)

Filling in x=0 gives y = 50 -20.5 = 29.5

So, the nearest of the offered choices is ...

  f(x) = 34.59 +3.74x

Please note that no calculator is involved. We simply estimated the parameters of the regression line based on a couple of the data points. This estimate is sufficient to allow us to find the correct answer choice.

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2. If we look at a few other data points, we see that they have a generally upward trend that matches the above equation pretty well. That suggests a very high value of correlation, so the appropriate choices would be 0.93 or 0.96. Of course, the correlation is positive, matching the sign of the slope of the regression line. (Since the data points do not fall exactly on a line, the correlation is not 1.00. My assessment is they fall close enough to the line to rate a correlation of .96 instead of .93.)

At this point, the output of your calculator can confirm that the correlation coefficient is closer to 0.96. Mine says r = 0.9624.

  r = 0.96

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3. We have already determined by examining the data points that the correlation is "very strong." The correlation coefficient of 0.96 confirms this.

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4. Using the equation of part 1, we can find the number of study hours:

  68 = 34.59 +3.74x

  (68 -34.59)/3.74 = x ≈ 8.933

We expect a score of 68 is associated with about 8.93 study hours.

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Please note that we have used very little calculation to determine the answers to this question. The only requirement is a general understanding of what linear regression and correlation coefficients mean.

3 0
4 years ago
Determine whether the lines l1 and l2 given by the vector equations are parallel, perpendicular, or neither. L1: r(t) = (-2 + 4t
ss7ja [257]

Over which interval(s) is the function decreasing?

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3 years ago
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to
Zepler [3.9K]

Step-by-step explanation:

We know that,

Z=\dfrac{X-\mu}{\sigma}

Where,

X = raw score = 84

μ = mean = 68

σ = standard deviation = 10

Putting the values,

Z=\dfrac{84-68}{10}=1.6

The percent of his sales that were less than or equal to 84 inches, is

P(Z\leq 1.6)=0.9452=94.52\%\approx 94.5\%

Therefore, the remaining 100-94.5=5.5% were more than 84 inches.

4 0
3 years ago
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