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3 years ago

Find the value of each variable in the circle to the right. The dot represents the center of the circle.

Mathematics

1 answer:

MrMuchimi3 years ago

4 0

9514 1404 393

Answer:

a = 57.5

b = 90

c = 65

Step-by-step explanation:

The measure of an inscribed angle is half that of the arc it intercepts. The angle at 'a' intercepts an arc of 115°, so its measure is (115/2)° = 57.5°

The inscribed angle at 'b' is a semicircle, 180°, so the measure of 'b' is (180/2)° = 90°.

The arc 'c' is the remainder of a semicircle after 115° have been taken away. Its measure is 180° -115° = 65°.

(a, b, c) = (57.5, 90, 65)

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The total number of cars produced by TESLA each year is y = 200000(1.25)x , where x represents the number of years after 2020. h

Rudiy27

Answer:

y=1250000

Step-by-step explanation:

Multiply 200000 by 1.25.

y = 250000 ⋅5

Multiply 250000 by 5.

y=1250000

3 0

3 years ago

Read 2 more answers

Please help! Giving points :)!

velikii [3]

Answer:

Step-by-step explanation:

3/5 + 1/4 = 0.85

0.85 as a fraction is 17/20

So the answer is C

6 0

3 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually. What would the value of the car be in 20

Fed [463]

Answer: -$6,500

Step-by-step explanation:

Here we could , use the arithmetic progression where

T(2020 - 2010) = a + ( n - 1 )d

T10 = a + ( 10 - 1 )d --------------- 1

a = $25,000, n = 10 and d = 14% of $25,000 = $3,500 the common difference.

Note since it decreases the common difference d = -$3,500.

Now substitute for the values in the equation above.

T10 = 25,000 + 9 x -3,500

= $25,000 - $31,500

= -$6,500 (deficit )

4 0

3 years ago

You decide to put the Gamer table on sale for 20% off. You normally charge $210 for this table. What is the amount of the discou

kvasek [131]

First, you would multiply $210 by 20%
This is done by converting 20% to a decimal, .2

So, 210 • .2 = 42

Then to find the sale price, you would subtract that 42 that you just got from 210,

and you get $168 as the final sale price.

5 0

3 years ago