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2 years ago

What is the GCF of 45,85and90

1 answer:

4 0

45 = 1, 3, 5, 9, 15, 45
85 = 1, 5, 17, 85
90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

So, greatest common factor is 5

You might be interested in

The original price is $60.The sales price is $45. What is the percent of discount?

MariettaO [177]

Hey there!

To start, first subtract 45 from 60 to find the amount of money taken off from the discount:
60-45
=15

Because 15 dollars have been taken off the original price, you can find the percent of discount by setting up this proportion where x represents the percent of discount:
15/60=x/100

Now, cross multiply and solve for x:
1500=60x
x=25

Therefore, your discount would be 25%.

Hope this helps and have a marvelous day! :)

5 0

2 years ago

F(x)=2+3(2x+4)-8 is??

Vikki [24]

X = -6/5 I hope this helps

3 0

2 years ago

Determine how many solutions exist for this system of equations:

IrinaK [193]

Answer:

No Solution

Step-by-step explanation:

-3x+6=-3x-5

+5 +5

-3x+11=-3x

+3 +3

11=0

No Solution

4 0

2 years ago

The diagram shows the net of a juice box. The box is a rectangular prism.

Hoochie [10]

Answer:

269.6 square centimeters

Step-by-step explanation:

The correct question is

The diagram shows the net of a juice box. the box is a rectangular prism. what is the surface area of the juice box? 6.4 cm 4 cm 10.5 cm

we know that

The surface area of a rectangular prism is equal to

$SA=2B+PH$

where

B is the area of the base

P is the perimeter of the base

H is the height of the prism

we have

$L=6.4\ cm\\W=4\ cm\\H=10.5\ cm$

Find the area of the base B

$B=LW$

substitute the given values

$B=(6.4)(4)=25.6\ cm^2$

Find the perimeter of the base P

$P=2(L+W)$

substitute the given values

$P=2(6.4+4)=20.8\ cm$

Find the surface area

$SA=2B+PH$

substitute

$SA=2(25.6)+(20.8)10.5=269.6\ cm^2$

6 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago