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Anna007 [38]
3 years ago
5

What is the value of x to the nearest tenth?

Mathematics
1 answer:
baherus [9]3 years ago
3 0
Answer:
x
≈
16

Explanation:
Remember Sine of an angle is
opposite side
hypotenuse

So let the hypotenuse be
x
:
sin
(
47
°
)
=
opposite side
x

sin
(
47
°
)
=
12
x

To solve for
x
, divide
sin
(
47
°
)
by
12
:
x
=
12
sin
(
47
°
)

x
≈
16
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What’s 4 over 5 fraction simplified
Naya [18.7K]

Answer:

It's already simplified to its lowest term so....

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2 years ago
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Hello, Can someone please explain on how to do this? thank you!
Marrrta [24]
Steps:

1) determine the domain

2) determine the extreme limits of the function

3) determine critical points (where the derivative is zero)

4) determine the intercepts with the axis

5) do a table

6) put the data on a system of coordinates

7) graph: join the points with the best smooth curve

Solution:

1) domain

The logarithmic function is defined for positive real numbers, then you need to state x - 3 > 0

=> x > 3 <-------- domain

2) extreme limits of the function

Limit log (x - 3) when x → ∞ = ∞

Limit log (x - 3) when x → 3+ = - ∞ => the line x = 3 is a vertical asymptote

3) critical points

dy / dx = 0 => 1 / x - 3 which is never true, so there are not critical points (not relative maxima or minima)

4) determine the intercepts with the axis

x-intercept: y = 0 => log (x - 3) = 0 => x - 3 = 1 => x = 4

y-intercept: The function never intercepts the y-axis because x cannot not be 0.

5) do a table

 x                          y = log (x - 3)

limit x → 3+            - ∞

3.000000001        log (3.000000001 -3) = -9

3.0001                  log (3.0001 - 3) = - 4

3.1                       log (3.1 - 3) = - 1

4                          log (4 - 3) = 0

13                       log (13 - 3) = 1

103                     log (103 - 3) = 10

lim x → ∞             ∞

Now, with all that information you can graph the function: put the data on the coordinate system and join the points with a smooth curve.
7 0
3 years ago
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Which expressions is equivalent ?
agasfer [191]
<h3>3 Answers: Choice C, Choice D, Choice F</h3>

What these expressions have in common is that their exponents multiply to 16

  • (-4)*(-4) = 16 for choice C
  • (-2)*(-8) = 16 for choice D
  • 8*2 = 16 for choice F

The rule used here is (a^b)^c = a^(b*c). We multiply the exponents and keep the base the same.

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The ratio of boys to girls on the bus is 20/15
saveliy_v [14]

Answer:

Equivalent ratios

20 = 40 = 60 = 80 and so on....

15 = 30 = 45 = 60 and so on.....

Both include 60

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3 years ago
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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
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