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3 years ago

Express 0.001 in words.

Mathematics

1 answer:

Elan Coil [88]3 years ago

7 0

Hi there!

In word form, 0.001 in worded form is one-hundredth.

Not actually a hundred, like, 100, but hundredTH, the TH meaning AFTER the decimal place.

This is just like how 0.01 is one tenTH, not ten like 10.

I am just saying that numbers after the decimal point will always end in TH.

Hope this helps!

Message me if you need anything else! I'd be happy to answer more questions you may have! :D

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An attendant recorded the data in the table showing the types and colors of vehicles that parked in a parking lot. (Refer to the

DedPeter [7]

Answer:

<h2>There are 22% of chances that the next vehicle in the lot is a red car.</h2>

Step-by-step explanation:

The event is a red car vehicle. So, the experimental probability would be the ratio between the total number of red car vehicles and the total number of cars.

Notice that there are 22 red vehicles in total, and there are 100 cars in total. So, the ratio is

$P=\frac{22}{100} = \frac{11}{50} = 0.22$

Therefore, there are 22% of chances that the next vehicle in the lot is a red car.

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4 years ago

I was wondering on how to do this

Kay [80]

Answer:

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

You start driving north for 9 miles, turn right, and drive east for another 40 miles. At the end of driving, what is your straig

Vsevolod [243]

The straight line distance from the starting point is 41 miles.

Explanation:

Given:

Distance covered towards north, n = 9 miles

Distance covered towards east, e = 40 miles

Distance from the origin to the end, x = ?

If we imagine this, then the route forms a right angle triangle

where,

n is the height

e is the base

x is the hypotenuse

Using pythagoras theorm:

(x)² = (n)² + (e)²

(x)² = (9)² + (40)²

(x)² = 1681

x = 41 miles

Therefore, the straight line distance from the starting point is 41 miles.

5 0

3 years ago

Trevor and $826 on account during 2.5% annual interest for seven years.

photoshop1234 [79]

Trevor invested $ 4720

Solution:

Given that, Trevor earned $ 826 on account during 2.5% annual interest for seven years

We can simple interest formula

The formula for simple interest is given as:

$\text{ Simple Interest } = \frac{p \times n \times r}{100}$

Where,

"p" is the principal

"n" is the number of years

"r" is the rate of interest

Here given that,

Principal = p = ?

rate of interest = r = 2.5 %

number of years = n = 7

Simple interest earned = 826

Substituting the values in above formula,

$826 = \frac{p \times 7 \times 2.5}{100}\\\\82600 = p \times 7 \times 2.5\\\\82600 = 17.5p\\\\p = \frac{82600}{17.5}\\\\p = 4720$

Thus he invested $ 4720

7 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago