<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.
2x^2+8xy+8y^2
2(x^2+4xy+4y^2)
She would have $7160.86 left over in cash here's why.....
she starts off with $8,322.65 and spends $830.98 on rent $8,322.65-$830.98= $7,491.67
then spends $210.36 on food and another $120.45 on gas
$7,491.67-$210.36= $7,281.31 - $120.45= $7,160.86 <---- Left over $