<h2>
Answer with explanation:</h2>
Given : In a restaurant, the proportion of people who order coffee with their dinner is p.
Sample size : n= 144
x= 120

The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-
Null hypothesis :
[ it takes equality (=, ≤, ≥) ]
Alternative hypothesis :
[its exactly opposite of null hypothesis]
∵Alternative hypothesis is left tailed, so the test is a left tailed test.
Test statistic : 

Using z-vale table ,
Critical value for 0.05 significance ( left-tailed test)=-1.645
Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.
Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.
Its 1400 divided by 224 so it will be 6.25 times
A)
Let X be the food expenditure of a family.
Let the population mean of an expenditure of a family be.
Let the population standard deviation of an expenditure of a family be.
The proportion families spend more than $30 but less than $490 per month on food is,
X - 4 +6 -x
x - x = -2
0 = -2
No solution
Answer:
1 and -5
Step-by-step explanation:
![~~~~~x^3+3x^2-9x+5=0\\\\\implies x^3-x^2+4x^2-4x-5x+5=0\\\\\implies x^2(x-1)+4x(x-1)-5(x-1)=0\\\\\implies (x-1)(x^2 +4x -5) = 0\\\\\implies (x-1)(x^2+5x -x -5) = 0\\\\\implies (x-1)[x(x+5) -(x+5)] =0\\\\\implies (x-1)(x-1)(x+5)= 0\\\\\implies (x-1)^2(x+5) = 0\\\\\implies x = 1,~~ x = -5](https://tex.z-dn.net/?f=~~~~~x%5E3%2B3x%5E2-9x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E3-x%5E2%2B4x%5E2-4x-5x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E2%28x-1%29%2B4x%28x-1%29-5%28x-1%29%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%20%2B4x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%2B5x%20-x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5Bx%28x%2B5%29%20-%28x%2B5%29%5D%20%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x-1%29%28x%2B5%29%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5E2%28x%2B5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20x%20%3D%201%2C~~%20x%20%3D%20-5)