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Ulleksa [173]
3 years ago
12

Using the Angle Bjsector theorem solve for x. Pleade show how you got your answer!

Mathematics
1 answer:
ladessa [460]3 years ago
7 0
The segments above the bisector are proportional to the corresponding segments below.

(4x +1)/5 = 15/3
4x +1 = 25
x = (25 -1)/4 = 6

The value of x is 6.
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Santino pays $36.85 for eleven 10-count boxes of granola bars. How much does 1 box of granola bars cost?
Fantom [35]

Answer:

One box of granola bars costs $3.35.

Step-by-step explanation:

36.85/11=3.35

5 0
3 years ago
The ratio of male to female students in a junior high
nalin [4]

the total no of population=1650

13:17 first add the two numbers 13+17=30

then 13/30×y =715 , i make 13 at the numerator because it's males ratio

after that 30/13×13/30y =715 ×30/13

y =715×30/13

y =1650

after u find the total no it's simple to find the no of females

17/30×1650=935

8 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
Suppose that someone is offering to sell you raffle tickets. There are blue, green, yellow, and red tickets available.
Ghella [55]

Answer:

Step-by-step explanation:

↑

8 0
3 years ago
How can i Round this answer to the nearest tenth? <br><br> 11.8
soldier1979 [14.2K]

9514 1404 393

Answer:

  11.8

Step-by-step explanation:

The number has no digits to the right of the tenths place, so the number is correctly rounded as is.

  11.8

5 0
3 years ago
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