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2 years ago

Is 52L greater than 5,200mL?

Mathematics

2 answers:

choli [55]2 years ago

6 0

1L = 1,000mL

therfore

52L = 52 × 1,000mL = 52,000 mL

52L = 52,000mL !!!!

Vesnalui [34]2 years ago

5 0

5200mL → 520 cL → 52dL → 5.2L

Example:

If

1L ------ 1000mL
x -------- 5200mL

1000x = 5200
 $x = \frac{5200}{1000}$
x = 5,2

Therefore:
52L > 5200mL or 5.2L (True)

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A pair of fair dice is cast. what is the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one

WITCHER [35]

If a pair of fair dice is cast, the probability that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is given by:

$\frac{P(sum\, of\, 9\, \cup\,3)}{P(3)}$

The number of outcomes of sum of 9 where at last one is 3 is (3, 6) and (6, 3) = 2.

The number of outcomes of last one of the numbers falling uppermost is a 3 is (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3) and (6, 3) = 11.

Therefore, the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is 2 / 11.

5 0

2 years ago

A rectangular bird feeder costs $18.00. A cylindrical bird feeder costs $24.00. The expected cost to keep the rectangular bird f

zheka24 [161]

18+4=27\qo/89 so that answer must me 18+4=27\qo but carry the y

8 0

2 years ago

8.06

Lera25 [3.4K]

Answer:

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Probability Problems.

Based on the study that Shane conducted we can assume a couple of things. Firstly, we can see that a vast majority of shoppers are buying products they saw advertised, so we can assume that advertising a product directly affects and contributes to the sales of that particular product.

Secondly, we can tell from the study that the number 1 method for advertising products is the Internet. Since 70% of all the interviewed shoppers are buying a product they saw on the internet.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

6 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

Linda commutes a total of 54 miles to and from work every day. She needs a new car, and good gas mileage is important to her. Sh

Alchen [17]

The answer is 1 gallon.

Miles per gallon(mpg) is computed by dividing the distance traveled by the how many gallons used. So you can derive a formula for how many gallons you would use given the mpg. You will end up with:

$gallon = \frac{miles travelled}{miles per gallon}$

The problem asks for how many gallons of gas she will safe in a five-day work work week. So first you need to compute how many miles that would be.

54 miles/day x 5days = 270 miles

So in a five day work week, she will travel 270 miles.

Now to see how much gas she will save, compute how many gallons she will use up for each car, given the mpg of each and find the difference.

First model:30 mpg

$gallons = \frac{270miles}{30mpg}$
$gallons = 9g$

This means that with the first model, she will have used up 9 gallons in a 5-day work week.

Second model: 27 mpg
$g = \frac{270mi}{27mi/g}$
$g = 10g$

This means that with the second model, she will have used up 10g in a 5-day work week.

Now for the last bit. How much will she save? You can get that by getting the difference of how many gallons each car would have used up.

10gallons - 9gallons = 1g

So she would have saved 1 gallon of gas if she buys the first car instead of the second.

8 0

1 year ago

Read 2 more answers