Answer:
2:11
Step-by-step explanation:
6 cups rolled oats, 2 cups mixed nuts, 1/2 cup sesame seeds, 1 cup dried cranberries, 1 cup dried unsweetened coconut, 1/2 cup honey. What is the proportion of cups of mixed nuts to the total cups of granola? The ratio of cups of mixed nuts to the overall cups of granola is 2:11.
Solution
Rolled oats= 6 cups
Mixed nuts=2 cups
Sesame seeds=1/2 cup
Cranberries= 1 cup
Dried unsweetened coconuts=1 cup
Honey =1/2 cup
All these ingredients combine to create granola
Total cups of granola= Rolled oats + Mixed nuts + Sesame seeds + Cranberries + Dried unsweetened coconuts + Honey
=6 + 2 + 1/2 + 1 + 1 + 1/2
=11 cups
The ratio of cups of mixed nuts to cups of granola= mixed nuts: total granola
=2:11
Answer:
The graph that represents the coordinates (-3,-3) and (0,3)
Step-by-step explanation:
The provided images illustrate the choices available.
The equation:
y+3=2(x+3)
is formulated in point-slope fashion, which is:
y-y₁=m(x-x₁)
where (x₁, y₁) denotes a point on the line and m indicates its slope. This implies that (-3,-3) resides on the line. To find the y-intercept, we substitute x = 0 into the equation, resulting in:
y+3=2(0+3)
y+3 = 6
y = 6 - 3
y = 3
This confirms that point (0, 3) is indeed on the line.
Answer: 
Step-by-step explanation:
Given: The manager of a video game store discovered that 35 out of the 140 customers who preordered the latest baseball game canceled their orders one day prior to the game's launch.
The likelihood that two customers who preordered the newest golf game will both cancel their orders the day before the game is released

Therefore, the probability that two customers who preordered the newest golf game will both cancel their orders the day before the game is released is
.
There is an issue with the question. The correct equation is as follows: the formula establishes the relationship between Celsius and Fahrenheit temperature readings. Knowing the Fahrenheit value ''F'' allows us to calculate Celsius ''C'', defining a function C(F) dependent on F. For part (a), we affirm that C is indeed a function of F. For part (b), we seek the mathematical domain of this function, which is all real numbers since there are no restrictions. Thus, the domain is Dom (C) = ( - ∞, + ∞). In part (c), water remains in liquid form at normal atmospheric pressure within the temperature range of 0 to 100 Celsius, so the range is Rang (C) = (0,100). We then determine the domain corresponding to this range by solving for variable ''F''; for C = 0 and C = 100 we find the relevant values for F. Consequently, the domain pertaining to liquid water temperature is Dom (C) = (32,212). Finally, for part (d), we simply substitute to find C where F = 71, which yields C ≈ 21.67 °C.