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3 years ago

What’s the smallest out of 45, -45, and 90

Mathematics

2 answers:

-BARSIC- [3]3 years ago

8 0

Answer:

-45

Step-by-step explanation:

- 45 because 45 and 90 is positive and -45 is negative so the - 45 would be smaller

valkas [14]3 years ago

5 0

Answer:

90 if fractions but -45 anythin else

Step-by-step explanation:

You might be interested in

Type the correct answer in each box.

Mariulka [41]

Answer:

$(x-5)^2+(y+4)^2=100$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The distance between the center and any point that lie on the circle is equal to the radius

we have the points

(5,-4) and (-3,2)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$r=\sqrt{(2+4)^{2}+(-3-5)^{2}}$

$r=\sqrt{(6)^{2}+(-8)^{2}}$

$r=\sqrt{100}\ units$

$r=10\ units$

step 2

Find the equation of the circle

we know that

The equation of a circle in standard form is equal to

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center

r is the radius

we have

$(h,k)=(5,-4)\\r=10\ units$

substitute

$(x-5)^2+(y+4)^2=10^2$

$(x-5)^2+(y+4)^2=100$

6 0

3 years ago

Another one I need help in asap

Paha777 [63]

Answer:

Gold : total

35 : 103

Step-by-step explanation:

USA gold = 35

Total usa = 103

Gold : total

35 : 103

4 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

Dividing Mixed Numbers?

scoray [572]

Answer:

80/7 simplified its 11 3/7

Step-by-step explanation:

lmk if you need an explanation! :]

8 0

3 years ago

Read 2 more answers

(01.06 MC) Jonna is 4 years older than her brother. Which of the following expressions shows the sum of their ages if b represen

Lana71 [14]

Your equation will be: b+4

6 0

3 years ago