Answer:
Area of rectangle = l*b
11*9
99
I think it will help you.
F and U and C and K and then finally U
To solve this question, you¨ll need to use a table of the standardized Normal distribution, or you could use a graphing calculator. The answer will be the cumulative property which will be a number with decimals. However, this represents the students who had a lower score then her, You´ll need to subtract this from 1 in order to get the opposite number, which will be the kids who scored higher. Be sure to multiply the result by 100 to get a percentage. Your answer should be 0.806 hope this helped!
Answer:
80 gummy bears are in 4 servings.
Step-by-step explanation:
2 servings=40 gummy bears
3 servings=60 gummy bears
<u>40</u>+20=<u>60</u>
Add 20 gummy bears for every serving.
5 servings=100 gummy bears
100-20=<u>80 gummy bears are in 4 servings</u>
OR you can also do:
40÷2=20 gummy bears are in each serving
So, 4x20=<u>80 gummy bears are in 4 servings</u>
<h3>
Answers: x = -17 and x = 64</h3>
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Explanation
Consider three scenarios:
- A) The value of x is the smallest of the set (aka the min)
- B) The value of x is the largest of the set (aka the max)
- C) The value of x is neither the min, nor the max. So 8 < x < 39.
These scenarios cover all the possible cases of what x could be. It's either the min, the max, or somewhere in between the min and max.
--------------------
We'll start with scenario A.
If x is the min, then that must mean 39 is the max as it's the largest of the set {18, 36, 16, 39, 27, 8, 34}
The range is 56, so,
range = max - min
56 = 39 - x
56+x= 39
x = 39-56
x = -17 which is one possible answer
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If instead we go with scenario B, then x is the max and 8 is the min
range = max - min
56 = x - 8
56+8 = x
64 = x
x = 64 is the other possible answer
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Lastly, let's consider scenario C. If x is not the min or the max, then it's somewhere between the min 8 and max 39. in short, 8 < x < 39.
Note that range = max - min = 39-8 = 31 which is not the range of 56 that we want. So there's no way scenario C can be possible here.
