Answer:
30% of ice cream are chocolate
Step-by-step explanation:
45% = vanilla
1/4 = strawberry
(1/4)×100 = 25%
chocolate = 100%-45%-25%
= 30%
In this question , it is given that p represents cost of 1 container of popcorn . Therefore cost of 5 containers of popcorn = 5p and cost of 3 containers of popcorn = 3p .
And g represents cost of 1 container of granola bars. THerefore cost of 4 containers of granola bars = 4g and cost of 6 containers of granola bars = 6g .
According to the given question, the required linear equations are
5p+4g=42.50 , 3p+6g =34.50 .
And these are the the required system of linear equation .
Answer:
The domain of the function is all real numbers
and the range is all positive real numbers 
Step-by-step explanation:
We have the following function
and we want to find the domain and the range.
The function we have is an example of an exponential function
with b > 0 and b ≠ 1. This types of functions in general have the following properties:
- It is always greater than 0, and never crosses the x-axis
- Its domain is the set of real numbers
- Its Range is the Positive Real Numbers

The domain of a function is the specific set of values that the independent variable in a function can take on.
When determining domain it is more convenient to determine where the function would not exist.
This function has no undefined points nor domain constraints. Therefore the domain is
.
The range is the resulting values that the dependent variable can have as x varies throughout the domain. Therefore the range is
.
We can check our results with the graph of the function.
Answer:
110 liters of water
Step-by-step explanation:
Let
x ----> liters of water
f(x) ----> the percent of the sat in the new mixture
we have

For 
substitute in the equation

solve for x


Answer:
Hello some parts of your question is missing below is the missing part
c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces
Answer: A) 0.0099
B) 0.6796
C) 0.13956
Step-by-step explanation:
weight of Navel oranges evenly distributed
mean ( u ) = 8 ounces
std ( б )= 1.5
navel oranges = X
A ) percentage of oranges weighing more than 11.5 ounces
P( x > 11.5 ) = 
= P ( Z > 2.33 ) = 0.0099
= 0.9%
B) percentage of oranges weighing less than 8.7 ounces
P( x < 8.7 ) = 
= P ( Z < 0.4667 ) = 0.6796
= 67.96%
C ) probability of orange selected weighing between 6.2 and 7 ounces?
P ( 6.2 < X < 7 ) = 
= P ( -1.2 < Z < -0.66 )
= Ф ( -0.66 ) - Ф(-1.2) = 0.13956