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3 years ago

Question no. e. and f. please help me !!!!!

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

If you need help again get that app photomath it is amazing(not sponsered)

Verizon [17]3 years ago

4 0

Answer: see proofs below

<u>Step-by-step explanation:</u>

Proofs are like a puzzle that you have to unravel.

You can manipulate the left side to get the right side (LHS → RHS)

You can manipulate the right side to get the left side (RHS → LHS)

You will need to use Pythagorean Identities, Sum and Difference Identities, and/or Double Angle Identities when manipulating one of the sides.

Use the following Identities:

cos 2A = cos²A - sin²A

sin 2A = 2 cosA · sinA <em>Note that A can be substituted with 2A </em>

sin 3A = 3 sinA - 4sin³A

cos 3A = 4cos³A - 3cosA

***********************************************************************************

e) Prove: 4 sin A · cos³A - 4 cos A · sin³A = sin(4A)

LHS → RHS

Given: 4 sin A · cos³A - 4 cos A · sin³A

Factor: 4 sin A · cos x (cos²A - sin²A)

2(2 sin A · cos A)(cos²A - sin²A)

Double Angle Identity: 2(sin 2·A)(cos 2·A)

2 sin (2A) · cos (2A)

Double Angle Identity: sin (2·2A)

sin 4A

Proven: sin 4A = sin 4A $\checkmark$

*****************************************************************************************

f) Prove: 4 sin³A · cos 3A + 4 cos³A · sin 3A = 3 sin(4A)

LHS → RHS

Given: 4 sin³A · cos 3A + 4 cos³A · sin 3A

Triple Angle Identity: 4sin³A (4cos³A - 3 cosA) + 4cos³A (3sinA - 4sin³A)

Expand: 16sin³A · cos³A - 12sin³A · cosA + 12cos³A · sinA - 16sin³A · cos³A

Simplify: 12cos³A · sinA - 12sin³A · cosA

Expand: 6cos²A(2cosA · sinA) - 6sin²A(2cosA · sinA)

Double Angle Identity: 6cos²A(sin 2A) - 6sin²A(sin 2A)

Factor: 6sin 2A (cos²A - sin²A)

Double Angle Identity: 6sin 2A (cos 2A)

3(2sin 2A · cos 2A)

Simplify: 3sin (2·2A)

3 sin 4A

Proven: 3sin 4A = 3sin4A $\checkmark$

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valentinak56 [21]

Answer:

a) 0.0954

b) 0.9045

c) 0.1601

Step-by-step explanation:

We are given the following information:

We treat first-year students having computers as a success.

P(first-year students have computers) = 54.3% = 0.543

Then the number of first year students follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

a) P(None have computers)

$P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954$

b) P(At least one has a computer)

$P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045$

c) P(All have computers)

$P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601$

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Question no. e. and f. please help me !!!!!​

Question no. e. and f. please help me !!!!!