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vladimir2022 [97]
3 years ago
14

A boy mentions that none of the 21 kids in his third-grade class has had a birthday since school started 56 days previously. ass

ume that kids in the class are drawn from a population whose birthdays have the same probability on all days of the year. what is the probability that 21 kids in such a class would not yet have a birthday in 56 days?
Mathematics
2 answers:
Nuetrik [128]3 years ago
7 0
21/356 because there are 21 kids and 365 days in a year
Vanyuwa [196]3 years ago
3 0

Answer:

0.8465²¹ = 0.0302

Step-by-step explanation:

If we assume that kids in the class are drawn from a population whose birthdays have the same probability on all days of the year, we have:

To not have a birthday in those 56 days, a kid should have his birthday in the remaining 309 days (365 - 56). Therefore, the probability of not having a birthday in those 56 days would be p = 309/365 = 0.8465.

Now we need that not only one, but 21 kids don't have a birthday in 56 days, this would be that the 1st kid doesn't have his birthday in those 56 days AND the second kid doesn't have it either AND the third either.... and so on for the 21 kids

This would give us that the probability is

p = 0.8465 x 0.8465 x 0.8465 x 0.8465 x...... x 0.8465  = 0.8465²¹ = 0.0302

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Step-by-step explanation:

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How does the graph y=3^x compare to the graph y=3^-x
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Answer:

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Y = 3x^x (-∞, 0) and (∞, ∞).

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Step-by-step explanation:

The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".

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3 years ago
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yKpoI14uk [10]
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If the mean of a given data set is 56 and the standard deviation is 10
aleksley [76]

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Step-by-step explanation:

5 0
3 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
MatroZZZ [7]

Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.

Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

Answer: the mean weight of the 11-person team is 208\dfrac{10}{11}\ lb.

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3 years ago
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