Question: The graph of a second-order polynomial is shown below, and the intercepts with the axes are marked. Explain how you ca
n use the graph to write the polynomial as a product of linear factors (x-a)(x-b). Be sure to state the values of a and b.
NEED HELP ASAP
NEED HELP ASAP
1 answer:
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The graph looks like this, on the enclosed pic:
One feature is that it's periodic and torn (has cut-off points), meaning the domain is the same as in case of tan(x): x€R and x =/= π/2+πn.
The range equals the range of arcsin(x): -π/2<=y<=π/2 OR y€[-π/2;π/2]
Hope could understand and if it helped! :)
One feature is that it's periodic and torn (has cut-off points), meaning the domain is the same as in case of tan(x): x€R and x =/= π/2+πn.
The range equals the range of arcsin(x): -π/2<=y<=π/2 OR y€[-π/2;π/2]
Hope could understand and if it helped! :)
Answer:
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806
Step-by-step explanation:
Check for conditions
For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:
a) Independence : we assume that the random sample of 400 students each student is independent from the other.
b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.
c) np= 400*0.74= 296>10
n(1-p) = 400*(1-0.74)=104>10
So then we have all the conditions satisfied.
Solution to the problem
For this case we know that the distribution for the population proportion is given by:
So then:
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806