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3 years ago

given a is equal to 60 degree and 30 degree prove that sin a + b is equal to sin a cos b + cos a sin b

Mathematics

1 answer:

Feliz [49]3 years ago

8 0

Answer:

A = 60 and B = 30

Now, A + B = 60 + 30 = 90

Also, A – B = 60 – 30 = 30

sin (A + B) = sin 90 = 1

sin A cos B + cos A sin B = sin 60 cos 30 + cos 60 sin 30

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3 years ago

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3 years ago

A physical fitness association is including the mile run in its secondary school fitness test. The time for this event for boys

Ierofanga [76]

Answer:

A) 0.0107

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 440 seconds and a standard deviation of 40 seconds.

This means that $\mu = 440, \sigma = 40$

Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.

This is the p-value of Z when X = 348. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{348 - 440}{40}$

$Z = -2.3$

$Z = -2.3$ has a p-value of 0.0107, and thus, the correct answer is given by option A.

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3 years ago

An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B. Eight random graduates from c

Karolina [17]

Answer:

Calculated value t = 0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

Step-by-step explanation:

we will t-test t = x⁻- y⁻ /S√ n₁+n₂-2

Given An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B

course A type at 62, 85, 59, 64, 73, 70, 75, and 72

mean of x is x⁻ = ∑x / n

mean of x = 62+85+59+64+73+70+75+ 72 /8 = 70

x⁻ = 70

course B type at 75, 64, 81, 55, 69, and 58

mean of y is y⁻ = ∑y / n = 75+64+81+55+69+ 58/6 = 67

y⁻ =67

course A course B

x y x- x⁻ (x-x⁻ )^2 y- y⁻ (y-y⁻ )^2

62 75 8 64 8 64

85 64 15 225 -3 9

59 81 -11 121 14 196

64 55 -6 36 -12 144

73 69 3 9 2 4

70 58 0 0 -9 81

75 5 25

72 2 4

∑( (x-x⁻ )^2= 484 ∑( (y-y⁻ )^2= 498

Adding ∑( (x-x⁻ )^2+ ∑( (y-y⁻ )^2= 982

n₁+n₂-2 = 8+6-2=12

now S^2 = 982/ 12 =81.833

Null hypothesis :H₀:μ₁=μ₂ ( there is no difference between A and B)

Alternative hypothesis :H₁:μ₁≠μ₂

level of significance ∝=0.05

we will use t-test t = 0.614 ( see attachment )

The degrees of freedom = n₁+n₂-2 = 8+6-2=12

From tabulated value t = 1.782

Calculated value t = 0.614< 1.782

we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .

There is no difference between Course A and Course B

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3 years ago

a pair of snow boots at an equipment store in big bear that originally cost $60 is on sale for $36.what is the rate of discount

Ksivusya [100]

The answer is a 40% discount because it costs $24 less than the original price and 24 divided by 60 is .4 or 40%!

4 0

3 years ago

Read 2 more answers

given a is equal to 60 degree and 30 degree prove that sin a + b is equal to sin a cos b + cos a sin b ​

given a is equal to 60 degree and 30 degree prove that sin a + b is equal to sin a cos b + cos a sin b