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3 years ago

ANSWER FAST: What is 14/15 hours ahead of 7:00 PM?

Mathematics

2 answers:

melomori [17]3 years ago

8 0

Answer:

8:00 to 9:00am

Step-by-step explanation:

7 plus 14=21 21-12= 9 am so between 9:00 and 10:00 am

kvasek [131]3 years ago

3 0

Answer:

Fixed answer: 7:56 PM

9:00AM/10:00AM

Step-by-step explanation:

Adding 12 hours to the current time has the effect of flipping AM to PM/PM to AM. Since 14 = 12 + 2 and 15 = 12 + 3, we can read "14/15 hours ahead" as an AM/PM flip, followed by adding 3/4 more hours. Following that here:

7:00 PM → 7:00 AM

7:00 AM + 2 hours/7:00 AM + 3 hours → 9:00 AM/10:00 AM

<h3>Fixed Response: 14/15 as a fraction</h3>

When I first answered this, I read 14/15 as "14 or 15 hours" instead of "14 fifteenths of an hour." Here's the explanation for that way of looking at it:

If we want to add 14 fifteenths of an hour to 7:00 PM, we first need to find what that is. Since there are 15 fifteenths of an hour in a whole hour, we can find 14 fifteenths by starting with one hour - 60 minutes - and taking away 1 fifteenth of an hour. 1/15 of 60 minutes = 60 ÷ 15 = 4 minutes, since 15 × 4 = 60. Subtracting those 4 minutes from 60, we find that 14 fifteenths of an hour = 60 - 4 = 56 minutes, and adding that to 7:00 PM gets us to 7:56 PM.

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Step-by-step explanation:

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4 years ago

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3 years ago

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Step-by-step explanation:

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enyata [817]

The correct option will be: C) 0.84

<u><em>Explanation</em></u>

Formula for Correlation coefficient :

$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}}$

First, for each point(x, y), we need to calculate x², y² and xy .

Then, we will find sum all x, y, x², y² and xy, which gives us Σx, Σy, Σx², ∑y² and Σxy

<em>(Please refer to the attached image for the table )</em>

Here we got, ∑x = 44 , ∑y = 183 , ∑x² = 362 , ∑y² = 6575 and ∑xy = 1480

'n' is the total number of data set, which is 7 here.

So, plugging those values into the above formula..........

$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}}\\ \\ r=\frac{7(1480)-(44)(183)}{\sqrt{[7(362)-(44)^2][7(6575)-(183)^2]}}\\ \\ r=\frac{10360-8052}{\sqrt{(598)(12536)}}\\ \\ r=\frac{2308}{\sqrt{7496528}}\\ \\ r=0.84$

So, the value of the correlation coefficient is 0.84

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3 years ago

Read 2 more answers