answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Solnce55 [7]
2 years ago
12

A survey asks 48 randomly chosen students if they plan to buy a school newspaper this week. Of the 48 surveyed, 32 plan to buy a

school newspaper. If 360 students bought papers, predict the number of students Enrolled at the school.
Mathematics
1 answer:
disa [49]2 years ago
8 0
The most probable number, in this case, would be by ratio,
32:48 = 360:x
cross multiply and solve for x
x=360*48/32=540
Ans. Most probably there are 540 students enrolled.
You might be interested in
Dapper Dan's Dancing School charges $85 for enrollment plus $6 per class, whereas Leaping Larry's Dancing School charges $40 for
gogolik [260]

Answer:

The Leaping Larry's Dancing School will be more expensive for a number of classes above 7.5 or equal to and above 8 classes

Step-by-step explanation:

The given information are;

The amount Dapper Dan's Dancing School charges for enrollment = $85

The amount Dapper Dan's Dancing School charges per class = $6

The amount Leaping Larry's Dancing School charges for enrollment = $40

The amount Leaping Larry's Dancing School charges per class = $12

Therefore, we have for X number of classes, when the cost of Dapper Dan's Dancing School and Leaping Larry's Dancing School are equal

$85 + X × $6 =  $40 + X × $12

$85 - $40 = X × $12 - X × $6

$45 =  X × $6

X = $45/$6 = 7.5

From which we have at X = 7.5 classes, the cost of the two dancing schools will be equal

However below 7.5 classes, for 7 classes for example, we have;

Cost of Dapper Dan's Dancing School = $85 + 7 × $6 = $127

Cost of Leaping Larry's Dancing School = $40 + 7 × $12 = $124

Therefore, for the number of classes below 7.5, Dapper Dan's Dancing School is more expensive than Leaping Larry's Dancing School

Above 7.5 classes, for 8 classes for example, we have;

Cost of Dapper Dan's Dancing School = $85 + 8 × $6 = $133

Cost of Leaping Larry's Dancing School = $40 + 8 × $12 = $136

Therefore, for the number of classes below 7.5, Dapper Dan's Dancing School is less expensive than Leaping Larry's Dancing School

Therefore, Leaping Larry's Dancing School will be more expensive for a number of classes above 7.5 or equal to and above 8 classes.

5 0
2 years ago
Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway be
exis [7]

Answer:

a. k = 3

b. Cumulative distribution function X, F(x)=\left \{ {0} , x\leq 1  \atop {1-x^{-3}, x>1}} \right.

c.  Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e.  Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

\int\limits^{\infty}_{-\infty} f(x)dx=1

So, the equation becomes,

\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3

b. For this problem, the cumulative distribution function is defined as :

F(x) = \int\limits^1_{\infty} f(x)dx +  \int\limits^x_1 f(x)dx

Now,

F(x) = 0 +  \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 +  \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}

Therefore the cumulative distribution function X is,

F(x)=\left \{ {0} , x\leq 1  \atop {1-x^{-3}, x>1}} \right.

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

P(2

≅ 0.088

d. Mean value of headway,

E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5

And,

E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3

The standard deviation of headway is,

= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

P(\alpha - \beta  < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X

From part b, F(x) = 0, if x ≤ 1

=1-(2.366)^{-3}\\=0.9245

8 0
2 years ago
This confuses me can I some help? Mary's mom gave her $100 to go shopping. She bought a shirt for $22.65 and a skirt $33.31. She
vazorg [7]
I believe the answer would be 49.64 because if you add them together 22.65+33.31 = 55.96 x .10 = 5.60 55.96-5.60 = $49.64
4 0
2 years ago
Read 2 more answers
Jillian has three different bracelets (x y and z) to give to her friends as gifts In any order she prefers if the bracelet y is
9966 [12]

Answer:

Number of ways to chose bracelet = 2 ways

Step-by-step explanation:

Given:

Total number of bracelet = 3

Y is chosen first

Find:

Number of ways to chose bracelet

Computation:

Y is chosen first so remain number of bracelet is 2

So,

Number of ways to chose bracelet = !2

Number of ways to chose bracelet = 2 × 1

Number of ways to chose bracelet = 2 ways

8 0
2 years ago
The average rate of change from x = 0 to x = 15 is about −4.667. How does the average rate of change from x = 0 to x = 20 compar
Amanda [17]
First we calculate the ambio rate from x = 0 to x = 15:
 We have:
 R = ((80-10) / (0-15))
 R = -4.666666667
 We now calculate for the entire interval:
 from x = 0 to x = 20:
 R = ((80-2) / (0-20))
 R = -3.9
 Answer:
 What is the average rate of change over the entire slide?
 R = -3.9
 The average rate of change from x = 0 to x = 15 is about -4.667. How does the average rate change from x = 0 to x = 20 compare to this number?
 The percentage difference is:
 (-3.9 / -4.666666667) * (100) = 83.57142857%
 100-83.57142857 = 16.42857143%
6 0
2 years ago
Other questions:
  • Javier uses 1/8 teaspoons of nutmeg for bread how much nutmeg do u need for 8 loafs of bread
    13·1 answer
  • Belkis is pulling a toy by exerting a force of 1.5 newtons on a string attached to the toy.
    15·2 answers
  • Livia’s bill for lunch came to $8 before tax and tip. The restaurant included a 7 percent sales tax, and she left a 15 percent t
    13·2 answers
  • If 6 components are drawn at random from the container, the probability that at least 4 are not defective is . If 8 components a
    6·1 answer
  • A store has a $120 dress. Then there is a 200% increase in the price of the dress. What is the final price of the dress?
    6·1 answer
  • Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent
    9·1 answer
  • Madden and Jenn both worked at the coffee shop today. Madden's total cups of coffee made is represented by f(x); and Jenn's tota
    8·2 answers
  • Carolyn
    12·2 answers
  • Phil sold pizza at the carnival for $1.75 a slice or $9.50 for a whole pizza he sold 76 slices of pizza and 64 whole pizzas he g
    9·1 answer
  • Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!