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Tcecarenko [31]

3 years ago

13

Find the perimeter and area

1 answer:

blondinia [14]3 years ago

7 0

Answer:

perimeter = 67

Step-by-step explanation:

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In PQR, PQ = QR. If M < P = 42, find the measures of the other angles

mihalych1998 [28]

Answer:

96

Step-by-step explanation:

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Colton was given a box of assorted chocolates for his birthday. Each night, Colton treated himself to some chocolates. There wer

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wait, im confused lol

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Please help, I need an explanation because I'm confused on how to do this.

gladu [14]

Answer:

15°

Step-by-step explanation:

If CB is a diameter, the arc BKAC is a semicircle and has a degree measure of 180. If the measure of angle BCK is 20 and it is an inscribed angle, then the measure of the arc it intercepts is twice that. So arc BK measures 40 degrees. That means that arc KA has a measure that is found by subtracting arc AC and arc BK from 180. 180 - 40 - 110 = 30. If that arc is 30 and the angle that intercepts it is inscribed, then the angle measure is half the measure of the arc. So the angle is 15°

3 0

4 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

Use the distributive property to simplify 2-(-3a+4)

dolphi86 [110]

Answer:

3a -2

Step-by-step explanation:

2-(-3a+4)

Distribute the minus sign to each term inside the parentheses

2 - -3a -4

2 + 3a -4

Combine like terms

2-4 +3a

-2 +3a

We usually put the variables first

3a -2

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