<h2>
Answer with explanation:</h2>
Given : Sample size 
Sample mean : 
Sample standard deviation : 
Let 
be the population mean of "like" ratings of male dates made by the female dates.
As per question ,
Null hypothesis : 
Alternative hypothesis : 
, It means the test is a one-tailed t-test. ( we use t-test when population standard deviation is unknown.)
Test statistic:
For 0.05 significance and df =188 (df=n-1) p-value = .001582. [By t-table]
Since .001582< 0.05
Decision: p-value < significance level , that means there is statistical significance, so we reject the null hypothesis.
Conclusion : We support the claim at 5% significance that t the population mean of such ratings is less than 8.00.
7 and 10 is half
so we expect the people come to wedding half of 120÷2=60 people
i just try
Answer:
(f-g)(x)=-x^(2)+2x+8
the solutions are:
<em><u>x=4 or x=-2</u></em>
Step-by-step explanation:
(f-g)(x)=2x+1-(x^(2)-7)
(f-g)(x)=-x^(2)+2x+1+7
(f-g)(x)=-x^(2)+2x+8
does this help or should I solve for the zeros/solutions of this quadratic equation?
then:
-x^(2)+2x+8=0
-(x^(2)-2x-8)=0
x^(2)-2x-8=0
(x-4)(x+2)=0
<em><u>x=4 or x=-2</u></em>
