Given that the score of the student on the quiz was 25c - 10(15 - c)
It can be seen that the first term of expression for her score (i.e. 25c) shows that 25 marks were awarded to each correct answer.
The second term of the expression (i.e. 10(15 - c)) represent that 10 marks were deducted from each wrong answer.
Thus, the number of wrong answers is 15 - c.
The sum of the correct answers and the wrong answers gives the total number of questions in the quiz.
Thus, the total number of questions in the quiz is given by 15 - c + c = 15.
Therefore, there are 15 questions in the quiz.
F(5) is equal to 17.
f(5)= 3(5)+2= 17
Answer:one solution
Step-by-step explanation:
Answer:
D. If the P-value for a particular test statistic is 0.33, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true.
D. Since this event is not unusual, she will not reject the null hypothesis.
Step-by-step explanation:
Hello!
You have the following hypothesis:
H₀: ρ = 0.4
H₁: ρ < 0.4
Calculated p-value: 0.33
Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In this case, you have a 33% chance of getting a value as extreme as the statistic value if the null hypothesis is true. In other words, you would expect results as extreme as the calculated statistic in 33 about 100 samples if the null hypothesis is true.
You didn't exactly specify a level of significance for the test, so, I'll use the most common one to make a decision: α: 0.05
Remember:
If p-value ≤ α, then you reject the null hypothesis.
If p-value > α, then you do not reject the null hypothesis.
Since 0.33 > 0.05 then I'll support the null hypothesis.
I hope it helps!
Answer:
Step-by-step explanation:
consider 6x^2-8x +k
disc .=(-8)^2-4*6*k=64-24k
it has no solution if 64-2k<0
or 64<2k
or 2k>64
k>32
it has two zeros if 64-2k≥0
or 64≥ 2k
or 2k≤64
or k≤ 32
it has one zero if Disc. =0
64-2k=0
2k=64
k=32
