Answer:
D=-4<0, the equation has no real solutions, it has two imaginary solutions.
Step-by-step explanation:
The quadratic equation
has coefficients
a=1,
b=4,
c=5.
The discriminant D is the expression

Hence,

Since the discriminant is negative, the equation has no real solutions, it has two imaginary solutions.
Answer:
Yes, there is evidence to support that claim that instructor 1 is more effective than instructor 2
Step-by-step explanation:
We can conduct a hypothesis test for the difference of 2 proportions. If there is no difference in instructor quality, then the difference in proportions will be zero. That makes the null hypothesis
H0: p1 - p2 = 0
The question is asking whether instructor 1 is more effective, so if he is, his proportion will be larger than instructor 2, so the difference would result in a positive number. This makes the alternate hypothesis
Ha: p1 - p2 > 0
This is a right tailed test (the > or < sign always point to the critical region like an arrowhead)
We will use a significance level of 95% to conduct our test. This makes the critical values for our test statistic: z > 1.645.
If our test statistic falls in this region, we will reject the null hypothesis.
<u>See the attached photo for the hypothesis test and conclusion</u>
Answer:
1 4/7
Step-by-step explanation:
11/7
7/7=1
11-7=4
1 4/7
Step-by-step explanation:
4 times = 4x increased by 7 = 4x+7

Answer:
Class interval 10-19 20-29 30-39 40-49 50-59
cumulative frequency 10 24 41 48 50
cumulative relative frequency 0.2 0.48 0.82 0.96 1
Step-by-step explanation:
1.
We are given the frequency of each class interval and we have to find the respective cumulative frequency and cumulative relative frequency.
Cumulative frequency
10
10+14=24
14+17=41
41+7=48
48+2=50
sum of frequencies is 50 so the relative frequency is f/50.
Relative frequency
10/50=0.2
14/50=0.28
17/50=0.34
7/50=0.14
2/50=0.04
Cumulative relative frequency
0.2
0.2+0.28=0.48
0.48+0.34=0.82
0.82+0.14=0.96
0.96+0.04=1
The cumulative relative frequency is calculated using relative frequency.
Relative frequency is calculated by dividing the respective frequency to the sum of frequency.
The cumulative frequency is calculated by adding the frequency of respective class to the sum of frequencies of previous classes.
The cumulative relative frequency is calculated by adding the relative frequency of respective class to the sum of relative frequencies of previous classes.