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3 years ago

How many measurements are less than 2 2/4 inches?

Mathematics

1 answer:

patriot [66]3 years ago

8 0

Centimeter
Milimeter
Micrometer
Nanometer

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Q is equidistant from the sides of ZTSR. Find mZRST. The

melisa1 [442]

9514 1404 393

Answer:

40°

Step-by-step explanation:

Triangles QRS and QTS are congruent (HL), so the marked angles are also congruent:

3x +2 = 4x -4

6 = x

Then the total angle measure of angle RST is ...

(3x +2) +(4x -4) = 7x -2 = 7(6) -2 = 40 . . . degrees

m∠RST = 40°

6 0

3 years ago

I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value o

fomenos

Answer:

The expected values from the product of 2 marbles are 1,2,3,4,5,6,8,10,12,15,20

Step-by-step explanation:

THe expected values are 1,2,3,4,5,6,8,10,12,15,20

4 0

3 years ago

A bag contains 5 red balls and 6 green balls. You plan to select 2 balls at random. Determine the probability of selecting 2 gre

Umnica [9.8K]

Step-by-step explanation:

first pick has probability

$\frac{6}{11}$

second pick has probability

$\frac{5}{10}$

combined probability is

$\frac{30}{110} = \frac{3}{11}$

7 0

3 years ago

A tank 4/5 full needs 75 litres to fill it up.how many litres does it hold when full

Ann [662]

X - the volume of the tank

$\frac{4}{5}x+75=x \\
75=x-\frac{4}{5}x \\
75=\frac{1}{5}x \ \ \ |\times 5 \\
x=375$

The tank holds 375 liters when full.

8 0

3 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago