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2 years ago

What are the sine and cosine ratios for a 30°–60°–90° triangle with a hypotenuse of 6, when θ = 60°?

Mathematics

1 answer:

Margaret [11]2 years ago

4 0

Answer:

There are two ways to do this problem algebraically or trigonometrically.

Algebraically/geometrically

The ratios of the sides of a 30/60/90 tri. are x, x√3, 2x (short leg, long leg, hyp). Therefore, if the long leg is 6 'units'. then 6 = x√3. x = 6√3.

The hyp is 2x then the hypotenuse is 2(6√3) = 12√3, rationalizing is 12√3/3 = 4√3

Using Trig.

We can use sinx = y/r = opp/hyp. The long leg of 6 is opposite 60 degrees (pi/3).

Therefore, sin(pi/3) = 6/r =

r = 6/sin(pi/3) = 6/(√3/2) = 12/√3, when you rationalize you get 12√3/3 = 4√3

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3 years ago

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3 years ago

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1 year ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago