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Leto [7]
3 years ago
14

it rained nine days in the month of april .based on this ,what is the probaility that is does not rain on the first day of may?​

Mathematics
1 answer:
attashe74 [19]3 years ago
6 0
<h3>Answer:     7/10</h3>

==========================================================

Explanation:

There are 30 days in April. Since it rained 9 of those days, the empirical probability of it raining in April is 9/30 = (3*3)/(3*10) = 3/10.

If we assume that the same conditions (ie weather patterns) hold for May, then the empirical probability of it raining in May is also 3/10. By "raining in May", I mean specifically raining on a certain day of that month.

The empirical probability of it not raining on the first of May is therefore...

1 - (probability it rains)

1 - (3/10)

(10/10) - (3/10)

(10-3)/10

7/10

We can think of it like if we had a 10 day period, and 3 of those days it rains while the remaining 7 it does not rain.

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Answer:

7 melons

Step-by-step explanation:

So we first figure out the unit price of each melon by dividing 7.5 with 3 and you get $2.5 per melon. Then you divide 17.5 by 2.5 and then you should get 7.

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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the
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Answer:

The wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} East of North  with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the  starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector \overrightarrow{DD'} is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

|AB|=|OA-OB|

\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|

\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|

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|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01 miles, and

\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

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