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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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The answer you are looking for is 45
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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
iren [92.7K]

Answer:

(y+1)

Step-by-step explanation:

use quadratic formula.

you will get two values of y

y= 7/5 and y= -1

u can create factors using these values of y (making 0 subject)

like

for 7/5

7/5=y

7=5y

5y-7=0

zero becomes subject in such cases bcz there are no numbers left on that side

now for -1

-1=y

y+1=0

so your factors are 5y-7 and y+1

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3 years ago
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Evaluate Y=3x-5 when x=6
Harman [31]
7 + (-3x^2) for x = 0 7 + [-3(0)^2]; = 7 + -3(0); = 7 - 0; = 7
6 0
3 years ago
In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th
ehidna [41]

Question:

In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th birthday, 25% married by their 20th birthday, and 76% married by their 30th birthday. Based on these data, what is the probability that in a family with two daughters, the first and second daughter will be married by the following ages? (Enter your answers to four decimal places.) (a) 18 years of age (b) 20 years of age (c) 30 years of age

Answer:

A.) 0.0064

B.) 0.0625

C.) 0.5776

Step-by-step explanation:

Given the following :

Married by 18th birthday = 8% = 0.08

Married by 20th birthday = 25% = 0.25

Married by 30th birthday = 76% = 0.76

In a family with two(2) daughters :

P(First daughter will be married by 18) = 0.08

P(second daughter will be married by 18) = 0.08

P(1st and 2nd married by 18) = (0.08×0.08) = 0.0064

B.)

P(First daughter will be married by 20) = 0.25

P(second daughter will be married by 20) = 0.25

P(1st and 2nd married by 20) = (0.25×0.25) = 0.0625

C.)

P(First daughter will be married by 30) = 0.76

P(second daughter will be married by 30) = 0.76

P(1st and 2nd married by 30) = (0.76×0.76) = 0.5776

6 0
3 years ago
Please answer this correctly without making mistakes
faltersainse [42]

Answer:

so to get a third you divide it by 3

first convert it to fraction

so it is 26/3

so do 26/3 divided by 3

so we do keep switch flip

26/3*1/3

so answer is 26/9 or 2 8/9

Step-by-step explanation:

6 0
3 years ago
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