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ad-work [718]
1 year ago
6

A security alarm requires a four-digit code. The code can use the digits 0–9 and the digits cannot be repeated.

Mathematics
2 answers:
never [62]1 year ago
7 0

Answer:

The answer is <em>C</em>

Explanation: Given:A security alarm requires a four-digit code. The code can use the digits 0–9 and the digits cannot be repeated.  

there is only 2 numbers which are greater than 7 i.e. 8 and 9. ∴ there is 2 possibility for first place.

For the remaining 3 digits there is 9 possibilities (including 1 which would left after choosing 1 from first place )

No of ways for the alarm code beginning with a number greater than 7= 2 P 1 x 9 P 3

<em>Total ways of code with 4 digits= 10 P 4.</em> Hope this helps :)

Oduvanchick [21]1 year ago
7 0

Answer: The answer is C

Step-by-step explanation:

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D(x) = \frac{7}{3} \sin \frac{2\pi }{365}x + \frac{35}{3}

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x = the number of days after March 21

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If x = -80, then,

D(-80) = \frac{7}{3} \sin \frac{2\pi }{365}(-80) + \frac{35}{3}

            =\frac{7}{3} \sin (-\frac{160 \pi}{365}) + \frac{35}{3}

            =-\frac{7}{3} \sin (\frac{160 \pi}{365}) + \frac{35}{3}

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Therefore, the number of hours of daylight on January 1 are 9.38.

The number of days from march 21 to may 4 = 44

If x =44, then,

D(44) = \frac{7}{3} \sin \frac{2 \pi}{365}(44) + \frac{35}{3}

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Therefore, the number of hours of daylight on May 4 are 13.27.

The number of days from October 28 to may 4 = 221

If x = 221, then,

D(221) = \frac{7}{3} \sin \frac{2 \pi}{365}(221) + \frac{35}{3}

           =\frac{7}{3} \sin (\frac{442 \pi}{365}) + \frac{35}{3}

           =\frac{7}{3}(-0.6152)+\frac{35}{3}

           \approx 10.23

Therefore, the number of hours of daylight on October 28 are 10.23.

4 0
2 years ago
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