Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
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Answer: 1.39
Step-by-step explanation:
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The total cereal mixture is 45 kilograms
<em><u>Solution:</u></em>
Given that, cereal mixture contains Almond, Wheat, and Corn in the ratio of 3:5:7
Therefore,
Almond : Wheat : Corn = 3 : 5 : 7
Let the almond present be 3x
Let the wheat present be 5x
Let the corn present be 7x
Therefore,
Total cereal mixture = 3x + 5x + 7x
Total cereal mixture = 15x
<em><u>Given that mixture contains 9 kg of almond</u></em>
Almond = 9
We know that,
Almond present = 3x
3x = 9
x = 3
Therefore,
Total cereal mixture = 15x = 15(3) = 45
Thus total cereal mixture is 45 kilograms
15 /6÷ 5/6
when we divide fractions we do the reciprocal of the second number that is of
5/6 = 6/5 and then multiply it by the first number
15/6 x 6/5
ans) 3