Answer:
Milk
Step-by-step explanation:
The amount of milk, cottage cheese and strawberries used are given in the attached picture.
3/4 cup of milk
2/6 cup cottage cheese
8/12 cup strawberry
Converting the fractions to decimal :
3/4 cup of milk = 0.75
2/6 cup cottage cheese = 0.3333
8/12 cup strawberry = 0.66666
0.75 > 0.6666 > 0.3333
We can conclude that amount of milk is greatest, then strawberry and the least is cottage cheese.
Solution:
<u>Given:</u>
Supplementary angles are a pair of angles that sum up to 180°.
<u>It should be noted:</u>
- If ∠G and ∠H are a pair of supplementary angles, they both sum up to 180°.
Equation formed: ∠G + ∠H = 180
<u>Substitute the values into the equation.</u>
- ∠G + ∠H = 180
- => 65 + ∠H = 180
<u>Subtract 65 both sides.</u>
- => 65 - 65 + ∠H = 180 - 65
- => ∠H = 180 - 65 = 115°
Answer:
5
Step-by-step explanation:
Answer:
1/4
Step-by-step explanation:
Both dice have odd numbers 1, 3, 5. The chances of getting an odd number on each number is 3/6
The probability of getting odd numbers on both is:
3/6 ⋅ 3/6 = 9/36 = 1/4
Mean of the distribution = u = 222
Standard Deviation = s = 16
We have to find the probability that a value lies between 190 and 230.
First we need to convert these data values to z score.
![z= \frac{x-u}{s}](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Bx-u%7D%7Bs%7D%20)
For x = 190,
![z= \frac{190-222}{16}=-2](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B190-222%7D%7B16%7D%3D-2%20)
For x = 230
![z= \frac{230-222}{16}=0.5](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B230-222%7D%7B16%7D%3D0.5%20)
So, we have to find the percentage of values lying between z score of -2 and 0.5
P( -2 < z < 0.5) = P(0.5) - P(-2)
From standard z table, we can find and use these values.
P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687
Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution.