1answer.
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
salantis [7]
3 years ago
9

A random sample of students were asked which award they would most like to win (Olympic Gold, Academy Award, or Nobel Prize), an

d asked which portion of the SAT they performed better on, between the Math and the Verbal. The results are summarized in the table below. Use this table to answer the following question.
1. If P(M) is the proportion of students who did better on the Math portion of the SAT who prefer winning an Olympic Gold, and P(V) is the proportion of students who did better on the Verbal portion of the SAT who prefer winning an Olympic Gold. What is the standard error of the distribution of the difference of proportions?

Mathematics
1 answer:
Komok [63]3 years ago
3 0

Answer:

Well would of them just ask how and when they will get it?

Step-by-step explanation:

So sorry if this is wrong.

You might be interested in
The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in the manufacturing sector.
elena-14-01-66 [18.8K]

Answer:

(a) The probability that the sample average will be between $30.00 and $31.00 is 0.5539.

(b) The probability that the sample average will exceed $21.00 is 0.12924.

(c) The probability that the sample average will be less than $22.80 is 0.04006.

Step-by-step explanation:

We are given that the hourly wage was $30.67 for Switzerland, $20.20 for Japan, and $23.82 for the U.S.

Assume that in all three countries, the standard deviation of hourly labor rates is $4.00.

(a) Suppose 40 manufacturing workers are selected randomly from across Switzerland.

Let \bar X = <u>sample average wage</u>

The z score probability distribution for sample mean is given by;

                                Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean wage for Switzerland = $30.67

            \sigma = standard deviation = $4.00

            n = sample of workers selected from across Switzerland = 40

Now, the probability that the sample average will be between $30.00 and $31.00 is given by = P($30.00 < \bar X < $31.00)

        P($30.00 < \bar X < $31.00) = P(\bar X < $31.00) - P(\bar X \leq $30.00)

        P(\bar X < $31) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{31-30.67}{\frac{4}{\sqrt{40} } } ) = P(Z < 0.52) = 0.69847

        P(\bar X \leq $30) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{30-30.67}{\frac{4}{\sqrt{40} } } ) = P(Z \leq -1.06) = 1 - P(Z < 1.06)

                                                             = 1 - 0.85543 = 0.14457

<em>The above probability is calculated by looking at the value of x = 0.52 and x = 1.06 in the z table which has an area of 0.69847 and 0.85543 respectively.</em>

Therefore, P($30.00 < \bar X < $31.00) = 0.69847 - 0.14457 = <u>0.5539</u>

<u></u>

(b) Suppose 32 manufacturing workers are selected randomly from across Japan.

Let \bar X = <u>sample average wage</u>

The z score probability distribution for sample mean is given by;

                                Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean wage for Japan = $20.20

            \sigma = standard deviation = $4.00

            n = sample of workers selected from across Japan = 32

Now, the probability that the sample average will exceed $21.00 is given by = P(\bar X > $21.00)

        P(\bar X > $21) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{21-20.20}{\frac{4}{\sqrt{32} } } ) = P(Z > 1.13) = 1 - P(Z < 1.13)

                                                          = 1 - 0.87076 = <u>0.12924</u>

<em />

<em>The above probability is calculated by looking at the value of x = 1.13 in the z table which has an area of 0.87076.</em>

<em />

(c) Suppose 47 manufacturing workers are selected randomly from across United States.

Let \bar X = <u>sample average wage</u>

The z score probability distribution for sample mean is given by;

                                Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean wage for United States = $23.82

            \sigma = standard deviation = $4.00

            n = sample of workers selected from across United States = 47

Now, the probability that the sample average will be less than $22.80 is given by = P(\bar X < $22.80)

  P(\bar X < $22.80) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{22.80-23.82}{\frac{4}{\sqrt{47} } } ) = P(Z < -1.75) = 1 - P(Z \leq 1.75)

                                                               = 1 - 0.95994 = <u>0.04006</u>

<em />

<em>The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.</em>

3 0
3 years ago
How do i write an exponential equation with a vertical asymptote other that y=0
antiseptic1488 [7]
Lets say we have
P(x)/q(x)

vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes


so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes

the horizontal assymtote
when the degree of P(x)<q(x), then HA=0

when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9





ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2

horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
 hope I helped, read the whole thing then ask eusiton

6 0
3 years ago
What is the area of this figure?
nalin [4]

Answer:

<em>290.5 square miles</em>

Step-by-step explanation:

Consider splitting this figure into two mini rectangles and a triangle, each of given lengths;

Dimensions of Rectangle 1 - Height ; 5 mi, Length ; 7 mi,\\Area of Rectangle 1 - Height * Length = 5 mi * 7 mi = 35 mi^2\\\\Dimensions of Rectangle 2 - Height ; 18 mi, Length ; 8 mi,\\Area of Rectangle 2 - Height * Length = 23 mi * 8 mi = 184 mi^2\\\\Dimensions of Mini Right Triangle - Height ; 11 mi, Base ; 8 + 5 = 13 mi,\\Area of Mini Right Triangle - 1 / 2 * Base * Height = 1 / 2 * 13 mi * 11 mi = 71.5 mi^2,\\\\Area of Figure = 35 mi^2 + 184 mi^2 + 71.5 mi^2 = 290.5 square miles

<em>Solution; 290.5 square miles</em>

4 0
3 years ago
Please Need Help. Thank You :)<br><br><br>Please find the Volume of these cylinders:
vladimir2022 [97]

Answer:

the one for the finding the height is 3.3, the one with 7 as the height is 351.86, and the one with 3 as the height is 150.8

Step-by-step explanation:

h=V

πr2=41.5

π·22≈3.30247

V=πr2h=π·42·7≈351.85838

V=πr2h=π·42·3≈150.79645

4 0
3 years ago
You are facing South. You make an about face, then turn 90 degrees right. What direction are you facing?
Roman55 [17]

Answer:

west

Step-by-step explanation:

4 0
3 years ago
Other questions:
  • Using the graph, find the value of y when x = 7. (image down below)
    9·1 answer
  • F(x)=x^2+7x-30<br> Find the intercepts, domain and range
    12·1 answer
  • Match each equation to the situation it represents.
    15·1 answer
  • What is the sign of the product?
    9·1 answer
  • What is the coefficient of the variable in the expression 2 - 5x – 4 + 8?
    6·1 answer
  • Expand than simplify: -2(8p+9q)
    7·1 answer
  • Analyze the characteristics of the hypotenuse of a right triangle. (select all that apply)
    11·2 answers
  • What fraction with a denominator of 12 is equivalent to 1/4?
    7·1 answer
  • Which of the following best defines a variable interval schedule of reinforcement?
    9·1 answer
  • An airliner has 40 rows of seats. Each row has 2 seats on one side, 3 seats in the middle and 2 seats on the other side. How man
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!