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Olin [163]
3 years ago
13

In the diagram below, ΔABC is inscribed in circle P. The distances from the center of circle P to each side of the triangle are

shown.
Answer the following questions:
What is the length of the radius of the circle shown?
What is the length of AB and AC?

Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0
Supposing the sides with 6 and 8 is a right angle, you can create a new line from C and P, and find the length using the equation of a²+b²=c² or 6²+8²=c², with c equaling the radius of the circle.

After finding c, you will have to find the length from C to the midpoint of AC, using the same equation a²+b²=c². If both the lengths of C to the midpoint of AC, and A to the midpoint of AC are equal, you can do b+b to find the length of AC.

Using the same approach, you can find AB. Hope this makes sense, if not, I can clarify more.
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2 Fred and Ted each have a savings account. Fred's savings account is represented in the table below.
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Answer:

you have to divide all the numbers and see what it is

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I need to know 2 digit times
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3 years ago
Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One
Naily [24]

Answer:

The expected value of X is E(X)=\frac{2754}{73} \approx 37.73 and the variance of X is Var(X)=\frac{226192}{5329} \approx 42.45

The expected value of Y is E(Y)=\frac{73}{2} \approx 36.5 and the  variance of Y is Var(Y)=\frac{179}{4} \approx 44.75

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and  probability mass function p(x). The expected value, denoted by E(X) or \mu_x, is

E(X)=\sum_{x\in D} x\cdot p(x)

The probability mass function p_{X}(x) of X is given by

p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function p_{Y}(x) of Y is given by

p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}

The expected value of X is

E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)

E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73

The expected value of Y is

E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)

E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2

The variance of X is

E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)

E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}

Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45

The variance of Y is

E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)

E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377

Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75

8 0
3 years ago
Huilan's age is two times Thomas's age. The sum of their ages is <br> 78<br> . What is Thomas's age?
lisov135 [29]

Let x equal Thomas' age.  Then Huilan's age is 2x.  Set up an equation to model this situation:

x+2x=78

<em>*Combine like terms*</em>

3x=78

<em>*Divide both sides by 3*</em>

x=26


Hope this helps!!

4 0
3 years ago
Read 2 more answers
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Ksenya-84 [330]
10% CDS are other types of music
4 0
3 years ago
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