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2 years ago

7

From her eye, which stands 1.69 meters above the ground, Penelope measures the angle of elevation to the top of a prominent skys

craper to be 16∘. If she is standing at a horizontal distance of 200 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.

1 answer:

exis [7]2 years ago

7 0

Answer: 59.04

Step-by-step explanation:

tan 16= x/200

tan16/1=x/200

200 tan 16=x

x=57.349077...

x=57.349077...

Add 1.69, the distance from the ground to her eye:

57.349077...+1.69= 59.039077...

≈ The skyscraper is 59.04 meters tall.

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Question 2 Multiple Choice Worth 1 points)

Sati [7]

Comparing g(x) with f(x), you can see that the function f(x) is translated to the right by 6 units to produce g(x) which is equivalent to (x-6)²

<h3>Transformation of function</h3>

Transformation is a techniques use to change the position of an object on an xy-plane.

Given the parent function f(x) = x² and the function g(x) = x²-12x +36

Factorize g(x);

g(x) = x²-6x-6x+36

g(x)=x(x-6)-6(x-6)

Group the terms to have;

g(x) = (x-6)²

Comparing g(x) with f(x), you can see that the function f(x) is translated to the right by 6 units to produce g(x) which is equivalent to (x-6)²

Learn more on transformation here: brainly.com/question/4289712
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4 0

2 years ago

assume that the number of production defects is 28 and that 14 of these are classified as major defects, 8 are classified as des

Arlecino [84]

Answer: 5

Step-by-step explanation:

Let A denotes the number of major defects and B denotes the number of design defect.

By considering the given information, we have

$U=28\ ;\ n(A)=14\ ;\ n(B)=8\ \ ;\ n(A^c\cap B^c)=1$

Now, the number of major defects or design defects:

$n(A\cup B)=U-n(A^c\cap B^c)=28-11=17$

Also,

$n(A\cap B)=n(A)+n(B)-n(A\cup B)\\\\\Rightarrow\ n(A\cap B)=14+8-17=5$

Hence, the number of design defects were major=5

6 0

3 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

3 years ago

40 = 90^2/16 sinx cosx find x

lbvjy [14]

Answer:

<h2>x = 4.545</h2>

Step-by-step explanation:

Given the expression

$40=\frac{90^2}{16} sinxcosx\\\\Cross\ multiplying;\\\\16*40 = 90^2 sinxcosx\\\\640 = 90^2 sinxcosx\\\\\frac{640}{8100} = sinxcosx\\ from\ trig\ identity, sin2x = 2sinxcosx\\sinxcosx = sin2x/2\\$

$Hence, \ \frac{640}{8100} = \frac{sin2x}{2} \\\\\frac{2*640}{8100} = sin2x\\ \\\frac{1280}{8100}=sin2x\\ \\0.158 = sin2x\\\\2x = sin^{-1} 0.158\\\\2x = 9.09\\x = 9.09/2\\x =4.545$

5 0

3 years ago

I need to find the points for a graph asap by finding the asymptote!!!

Korvikt [17]

Answer:

y=0

Step-by-step explanation:

Exponential functions, always have a horizontal asymptote. A function of the form f(x) = a (b^x) + c always has a horizontal asymptote at y = c. In this case the function is

$y=3^{x+2} \\\\We\ can\ rewrite\ it\ like\\\\\y=3^x.3^2\\y=9(3^x)$

so it doesn't have a +c which means the horizontal asymptote is y = c = 0

y = 0 is the horizontal asymptote

8 0

2 years ago