In a class of 160 students, 90 are taking math, 78 are taking science, and 62 are taking both math and science. What is the prob
ability of randomly choosing a student who is not taking science?
69%
44%
51%
49%
69%
44%
51%
49%
2 answers:
Answer:
Option 3 - 51%
Step-by-step explanation:
Given : In a class of 160 students, 90 are taking math, 78 are taking science, and 62 are taking both math and science.
To find : What is the probability of randomly choosing a student who is not taking science?
Solution :
We can show this situation through Venn diagram,
Refer the attached figure below.
Take the Blue circle as the students taking math and Red circle as the student taking science.
Total number of student = 160
Let M be the student taking math M=90
S be the student taking math S=78
The student only take math is 90-62=28=P
The student only take science is 78-62=16=Q
Total students cover the circle is 28+62+16 = 106
Remaining students who are not in either two circles is 160-106=54
The remaining students and student taking math only is 28+54=82
So, 82 students are those who are not taking science.
The probability of student who is not taking science is
In percentage, 0.51=51%.
Therefore, Option 3 is correct.
The probability of student who is not taking science is 51%.
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Answer:
a
The Null hypothesis represented as
The Alternative hypothesis represented as
b
p-value = P(Z < -3.37 ) = 0.000376
c
There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students
Step-by-step explanation:
From the question we are told that
The population size is
The sample size for graduates is
The sample mean for graduates is
The sample standard deviation for graduates is
The sample size for under-graduates is
The sample mean for under-graduates is
The sample standard deviation for graduates is
The Null hypothesis represented as
The Alternative hypothesis represented as
Where are the population mean
Now the test statistic is mathematically represented as
substituting values
The p-values is mathematically evaluated as
p-value = P(Z < -3.37 ) = 0.000376
The above answer is gotten using a p-value calculator at (0.05) level of significance
Looking the p-value we see that it is less than the level of significance (0.05) so Null hypothesis is rejected
Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students
Answer:
The system of equations has no solution.
Step-by-step explanation:
Given the system of equations
2y = 2x+7
x = y+2
solving the system of equations
Arrange equation variables for elimination
Multiply -y+x=2 by 2: -2y+2x=4
so adding
so the system of equations becomes
0 = 11 is false, therefore the system of equations has no solution.
Thus,
No Solution!
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>Step-by-step explanation:
Given the function j(x) = 11.6 and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6 and k(x) =
are inverse functions.