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fgiga [73]
3 years ago
12

There are 102102 students in a statistics class. The instructor must choose two students at random. Students in a Statistics Cla

ss Academic Year Statistics majors non-Statistics majors Freshmen 1414 55 Sophomores 55 1616 Juniors 1818 66 Seniors 2020 1818 Copy Data What is the probability that a freshman non-Statistics major and then a junior Statistics major are chosen at random? Express your answer as a fraction or a decimal number rounded to four decimal places.
Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

0.0087 probability that a freshman non-Statistics major and then a junior Statistics major are chosen at random

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

What is the probability that a freshman non-Statistics major and then a junior Statistics major are chosen at random?

There are 5 freshman non-Statistics majors out of 102 students.

Then, there will be 18 junior statistics majors out of 101 students(1 will have already been chosen). So

p = \frac{5}{102}*\frac{18}{101} = \frac{5*18}{102*101} = 0.0087

0.0087 probability that a freshman non-Statistics major and then a junior Statistics major are chosen at random

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In-s [12.5K]

Answer:

x = 11

Step-by-step explanation:

sides are proportional so you can set up a proportional and cross-multiply:

JL is to PQ as LK is to QR

x/22 = 8/16

x/22 = 1/2

2x = 22

x = 11

5 0
3 years ago
Based on the survey results from 300 selected students in the Burbank School District (BSD), the 98% confidence interval used to
Pavlova-9 [17]

Answer:

follows are the solution to this question:

Step-by-step explanation:

Please find the correct question in the attached file:

The formula for calculating the Confidence interval of proportion:

\to CI=\hat{p} \pm Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\\to  \hat{p}=\frac{\text{lower limit +upper limit}}{2}\\

      =\frac{0.48+0.60}{2}\\\\=\frac{1.08}{2}\\\\=0.54

The number of learners with access to working at home on a computer:

= 0.54 \times 300 \\\\=162

Lower limit =0.48

upper limit = 0.60

\to margin \ error = \frac{\text{upper limit - lower limit}}{2}\\

                          =\frac{0.60-0.48}{2}\\\\=\frac{0.12}{2}\\\\= 0.06

5 0
3 years ago
What is an equation of the line that passes through the points (-5,6) and (7,6)?​
Tatiana [17]
The y values are not changing so y = 6 describes the line
3 0
3 years ago
Choose two positive integers a and b, making sure that a&lt;5 and b&gt;80 . The width of a rectangular computer screen is a inch
qwelly [4]

Answer:

Dimensions of the screen are a minimum of 10 inches wide by 8 inches high. The difference between the height and width will be 2 inches

Step-by-step explanation:

Height = H

Width = H + a

Area = Height x Width

b = H * (H + a)

b = H^{2} + aH

H^{2} + aH > 80

H^{2} + aH - 80 > 0

a<5 means 'a' can be a = 1, 2, 3, 4

Solving for H for each option of 'a' will give values of 'H' using the quadratic formula below

H = \frac{-b\±\sqrt{b^{2}-4ac}}{2a}

As 'a' and 'b' must be positive integers, H and W must be positive integers as well,

a = 1, H = 8.46, H = -9.46

a = 2, H = 8, H = -10

a = 3, H = 7.57, H = -10.57

a = 4, H = 7.17, H = -11.17

Based on above possible answers:

a=2, H=8

W = H + a

W = 8 + 2

W = 10

Dimensions of the screen are a minimum of 10 inches wide by 8 inches high. The difference between the height and width will be 2 inches

5 0
3 years ago
g (15 points) Suppose 42 out of 600 rats exposed to a potential carcinogen develop tumors. A control group of 350 rats not expos
vazorg [7]

Answer:

a) The relative risk is 1.8\overline{846153}

b) The attributable risk is 0.69 \overline{047619}

c) There is a relationship between exposure and tumor risk

Step-by-step explanation:

The number of exposed rats that develop tumors, a = 42

The number of rats exposed to the carcinogen = 600 rats

The number of exposed rats that did not develop tumors, b = 600 - 42 = 558

The number of not exposed rats in the control group = 350 rats

The number of rats that develop tumors in the control group, c = 13 tumors

The number of not exposed rats that did not develop tumors, d = 350 - 13 = 337  rats

a) The relative risk, RR = a/(a + b)/(c/(c + d))

∴ RR = (42/(42 + 558))/(13/(13 + 337)) = 49/26 = 1.8\overline{846153}

b) The attributable risk = (a - c)/a

∴ The attributable risk = (42 - 13)/42 = 0.69 \overline{047619}

c) The odds ratio = (a·b)/(c·d)

∴ The odds ratio = 42 × 558/(13 × 337) = 23439/4381 ≈ 5.35

Given that the result of attributable risk is positive, there is an indication that there is a higher probability to develop tumor when exposed to the potential carcinogen, therefore, there is a relationship between exposure and tumor risk

7 0
3 years ago
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