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Snowcat [4.5K]
3 years ago
6

A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells e

very 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together
Mathematics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

As few as just over 345 minutes (23×15) or as many as just under 375 minutes (25×15).

Imagine a simpler problem: the bell has rung just two times since Ms. Johnson went into her office. How long has Ms. Johnson been in her office? It could be almost as short as just 15 minutes (1×15), if Ms. Johnson went into her office just before the bell rang the first time, and the bell has just rung again for the second time.

Or it could be almost as long as 45 minutes (3×15), if Ms. Johnson went into her office just after the bells rang, and then 15 minutes later the bells rang for the first time, and then 15 minutes after that the bells rang for the second time, and now it’s been 15 minutes after that.

So if the bells have run two times since Ms. Johnson went into her office, she could have been there between 15 minutes and 45 minutes. The same logic applies to the case where the bells have rung 24 times—it could have been any duration between 345 and 375 minutes since the moment we started paying attention to the bells!

Step-by-step explanation:

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Answer:

-3<x<3

Step-by-step explanation:

let y=0

0=x^2-9

0=(x+3)(x-3)

x=-3 and x=+3

8 0
3 years ago
Y=30w+200 please help me
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Answer: That is the final answer, y=30w+200

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o
Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(XE(X_{k}) = n_{k} * p_{k}

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = \frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= \frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= \frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\ = 0.000001

c.

P(X₁≤2) = \frac{22!}{0!} * 0.08^{0} * (0.92)^{22} + \frac{22!}{1!} * 0.08^{1} * (0.92)^{21} + \frac{22!}{2!} * 0.08^{2} * (0.92)^{20} = 0.7442

I hope you have a SUPER day!

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The car is faster because it’s 120km/hr, wile the train is 75km/hr
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