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gladu [14]
3 years ago
8

Find the hypotenuse of each isosceles right triangle when the legs are of the given measure.

Mathematics
1 answer:
kow [346]3 years ago
8 0

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{3\sqrt{2}}\\ b=\stackrel{opposite}{3\sqrt{2}}\\ \end{cases} \\\\\\ c=\sqrt{(3\sqrt{2})^2+(3\sqrt{2})^2}\implies c=\sqrt{3^2\sqrt{2^2}+3^2\sqrt{2^2}}\implies c=\sqrt{18+18} \\\\\\ c=\sqrt{36}\implies c=6

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Angle 6= (11x+8) and angle 7=(12x-4) what is the measure of angle 4
Hitman42 [59]

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Answer is m∠4=40

Step-by-step explanation:

take note that m∠6 & m∠7 are vertical angles. Vertical angles are equal to each other, therefore m∠6 is equal to m∠7.

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Let X be the number of the cars being repaired at a repair shop. We have the following information:
worty [1.4K]

Answer:

(a) Sample Space

S = \{0,1,2,3\}

(b) PMF

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}

(c) CDF

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

S = \{0,1,2,3\}

Solving (b): The PMF

From the question, we have:

P(2) = P(1)

P(0) = P(3)

P(1\ or\ 2) = 0.5 * P(0\ or\ 3)

P(1\ or\ 2) = 0.5 * P(0\ or\ 3) can be represented as:

P(1) + P(2) = 0.5[P(0) + P(3)]

Substitute P(2) = P(1) and P(0) = P(3)

P(1) + P(1) = 0.5[P(0) + P(0)]

2P(1) = 0.5[2P(0)]

2P(1) = P(0)

P(0)= 2P(1)

Also note that:

P(0) + P(1) + P(2) + P(3) = 1

Substitute P(2) = P(1) and P(0) = P(3)

P(0) + P(1) + P(1) + P(0) = 1

2P(1) + 2P(0) = 1

Substitute P(0)= 2P(1)

2P(1) + 2*2P(1) = 1

2P(1) + 4P(1) = 1

6P(1) = 1

Solve for P(1)

P(1) = \frac{1}{6}

To calculate others, we have:

P(2) = P(1)

P(2) = P(1) = \frac{1}{6}

P(0)= 2P(1)

P(0) =2 * \frac{1}{6}P(0) =\frac{1}{3}

P(3) = P(0) =\frac{1}{3}

Hence, the PMF is:

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x}  P(x_i)

For x = 0;

We have:

P(X \le 0) = P(0)

P(X \le 0) = 1/3

For x = 1

P(X \le 1) = P(0) + P(1)

P(X \le 1) =  1/3 + 1/6

P(X \le 1) =  1/2

For x = 2

P(X \le 2) =  P(0) + P(1) + P(2)

P(X \le 2) =  1/3 + 1/6 + 1/6

P(X \le 2) =  2/3

For x = 3

P(X \le 3) =  P(0) + P(1) + P(2) + P(3)

P(X \le 3) =  1/3 + 1/6 + 1/6 + 1/3

P(X \le 3) =  1

Hence, the CDF is:

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}

<em>See attachment (2) for histogram</em>

6 0
3 years ago
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