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sattari [20]
3 years ago
11

Find sin 2x, cos 2x, and tan 2x from the given information. cscx=4,tanx<0.

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

Step-by-step explanation:

Given the expression cosec (x) = 4 and tan(x)< 0

since cosec x = 1/sinx

1/sinx = 4

sinx = 1/4

From SOH, CAH TOA

sinθ = opposite/hypotenuse

from sinx = 1/4

opposite = 1 and hypotenuse = 4

to get the adjacent, we will use the Pythagoras theorem

adj² = 4²-1²

adj² = 16-1

adj ²= 15

adj = √15

cosx = adj/hyp = √15/4

tanx = opposite/adjacent = 1/√15

since tan < 0, then tanx = -1/√15

From double angle formula;

sin2x = 2sinxcosx

sin2x = 2(1/4)(√15/4)

sin2x = 2√15/16

sin2x = √15/8

for cos2x;

cos2x = 1-2sin²x

cos2x = 1-2(1/4)²

cos2x = 1-2(1/16)

cos2x= 1-1/8

cos2x = 7/8

for tan2x;

tan2x  = tanx + tanx/1-tan²x

tan2x = 2tanx/1-tan²x

tan2x = 2(-1/√15)/1-(-1/√15)²

tan2x = (-2/√15)/(1-1/15)

tan2x = (-2/√15)/(14/15)

tan2x = -2/√15 * 15/14

tan2x = -30/14√15

tan2x = -30/7√15

rationalize

tan2x =  -30/7√15 * √15/√15

tan2x =  -30√15/7*15

tan2x =  -2√15/7

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The width of the window should be 7.6\ ft and the height of the window should be 3.8\ ft

Step-by-step explanation:

we know that

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4 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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3 years ago
By rounding to 1 significant figure work out 54×82​
pav-90 [236]

Answer:

4000

Step-by-step explanation:

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