All categories

2 years ago

A can contains 40 fluid Ounces of fruit juice. How many pints of fruit juice

Mathematics

2 answers:

expeople1 [14]2 years ago

7 0

Answer:

2.5 pints of fruit juice

vaieri [72.5K]2 years ago

3 0

Answer:

2.5

Step-by-step explanation:

You might be interested in

Im doing decimals greatest to least and i need help with number 12 and 13

erma4kov [3.2K]

(I know there isnt any real problem here to solve, but heres a tip on how to solve greatest to least with decimal problems)

1. Just because a number looks big, doesn't mean it is big

Example: 1.00000000000001 < 1.1

just look at the numbers ^ and dont just hastily read it over assuming that 1.1 is smaller because it "has less digits"

2. Negative numbers are... opposite. and they are less than positive numbers

-3.4 > -4.1

Why is this? Well, if you look on a line, with the point 0 in the middle, you can see that -3.4 is not as far away from 0 as -4.1 is. So the idea is to apply opposite logic for negative numbers

I hope these tips helped!! :D

3 0

2 years ago

Read 2 more answers

At the beginning of year 1, Zack invests $700 at an annual compound interest

REY [17]

Answer: the answer is C

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Simplify: 8/6/7 helpppppppppppp meeeeeeeeeeeee I will give you 100 points but only if you do right I will report if you play me

Naily [24]

Answer:

4/12??

Step-by-step explanation:

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

What is -6x plus 18 equals -6

Neporo4naja [7]

It’s going to be -18

First add 6 to both sides to get x by itself

-6x + 18 = -6
+6 +6

X + 18 = 0

Then minus the 18 giving you x = -18

7 0

2 years ago