Answer:
Inequality Form:
p > -3
Hope this helps!
Step-by-step explanation:
6p - 5p is 1p but you only but p and -3 stays there.
For # 3 and 5 you need to use the quadratic formula:
<span>(-b +/- srt(b^2 - 4ac))/2a </span>
<span>a, b, and c are representative of this formula: ax^2 +bx + c </span>
<span>1) 2x^2+3x-9=0 </span>
<span>(2x - 3)(x + 3) = 0 </span>
<span>2x - 3 = 0, x + 3 = 0 </span>
<span>+3 +3, -3 -3 </span>
<span>2x = 3, ***x = -3*** </span>
<span>/2 /2 </span>
<span>***x = 3/2*** </span>
<span>2) 5x^2+2x=0 </span>
<span>(x)(5x + 2) = 0 </span>
<span>5x +2 = 0, ***x = 0*** </span>
<span>-2 -2 </span>
<span>5x = -2 </span>
<span>/5 /5 </span>
<span>***x = -2/5*** </span>
<span>4) 4x^2+7x-2=0 </span>
<span>(4x - 1)(x + 2) = 0 </span>
<span>4x - 1 = 0, x + 2 = 0 </span>
<span>+1 +1, -2 -2 </span>
<span>4x = 1, ***x = -2*** </span>
<span>/4 /4 </span>
<span>***x = 1/4***</span>
32=-30
..................................
Answer: 0.9938
Step-by-step explanation:
Let x be the random variable that represents the daily revenue at a university snack bar.
As per given , we have
, 
and n= 100
Using formula 
,
z-score for x= 2600
The probability that the average daily revenue of the sample is higher than $2600 :
[P(Z>-z)=P(Z<z)]
Therefore, the probability that the average daily revenue of the sample is higher than $2600 = 0.9938
