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larisa86 [58]
2 years ago
8

Ranking the correct answer brainliest !

Mathematics
2 answers:
nikklg [1K]2 years ago
6 0

Answer:

angle 3 =68°+47°=115°

Hope it helps you

AnnZ [28]2 years ago
4 0

Answer:

∠ 3 = 115°

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ 3 is an exterior angle of the triangle, then

∠ 3 = ∠ 1 + ∠ 2 = 68° + 47° = 115°

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.. 5(x -1) = 2(2x +1) . . . . . . multiply by 20
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7 0
3 years ago
1. The US Mint has a specification that pennies have a mean weight of 2.5g. From a sample of 37 pennies, the mean weight is foun
Evgesh-ka [11]

Answer:

the pennies does not conform to the US mints specification

Step-by-step explanation:

z = (variate -mean)/ standard deviation

z= 2.5 - 2.4991 / 0.01648 = 0.0546

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checking for z= 0.0 under 55 gives 0.0219 (value gotten from the table of normal distribution)

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since 0.4781 > 0.05claim, therefore, the pennies does not conform to the US mints specification

the claim state a 5% significance level whereas the calculated significance level is 47.81%. therefore, the claim should be rejected

7 0
3 years ago
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The integral of (5x+8)/(x^2+3x+2) from 0 to 1
Gnom [1K]
Compute the definite integral:
 integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
 = integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:
 = 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.
 (5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand 1/(x^2 + 3 x + 2), complete the square:
 = (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx

For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds

Factor -1/4 from the denominator:
 = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds

Factor out constants:
 = (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds

Factor -1 from the denominator:
 = (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds

For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
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Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
 = (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5


Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
 = (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)

Which is equal to:

Answer:  = log(18)
5 0
3 years ago
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