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MAXImum [283]
3 years ago
14

Create three problems that require you to solve each using each of the trig ratios (one problem where you will use sine, one pro

blem where you will use tangent and one problem where you will use cosine) and then solve each problem, show all your work.

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0
See the attached image for the drawings of the problems. The figures are not to scale. The decimal values in each figure are approximate to one decimal place.

----------------------------------------------

Focus on Figure 1

sin(angle) = opposite/hypotenuse
sin(A) = BC/AC
sin(36) = 13.2/x
x*sin(36) = 13.2
x = 13.2/sin(36)
x = 13.2/0.58778525229248 <<-- make sure calc is in degree mode
x = 22.4571813404936
x = 22.5
This value is approximate (rounded to one decimal place)

----------------------------------------------

Move onto Figure 2

cos(angle) = adjacent/hypotenuse
cos(D) = DE/FD
cos(50) = y/57.4
57.4*cos(50) = y
y = 57.4*cos(50)
y = 57.4*0.64278760968653
y = 36.8960087960069
y = 36.9
This value is approximate (rounded to one decimal place)

----------------------------------------------

Finally, move onto Figure 3

tan(angle) = opposite/adjacent
tan(G) = JH/GH
tan(18) = z/10
10*tan(18) = z
z = 10*tan(18)
z = 10*0.3249196962329
z = 3.249196962329
z = 3.2
This value is approximate (rounded to one decimal place)

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Write the Equation of the conic.
tigry1 [53]

Answer:

(x² -1) + y² = 9

Step-by-step explanation:

diameter=6 so radius means 6/2=3 (thus,radius²=3²=9)

centre (1 and midpoint of diameter is at y=0, so 1,0)

3 0
2 years ago
Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the
Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

Z = \frac{X - \mu}{s}

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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Ricky can type 161 words in 7 minutes. how many can words can he type in 20 minutes.
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(2x-4)(x+5)  expand with indicated multiplication...

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