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3 years ago

The following data are from a completely randomized design.

1 answer:

5 0

Sorry I need the 2 answers to ask my own. I wouldn’t of understood this wnywyas. Hopefully you find someone who can do it

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Mark rode his bike 22/8 miles. which mixed number shows the fraction of miles he rode his bike

Vlad1618 [11]

Answer:

Hey there!

There aren't any options, but the correct answer would be 2 6/8, which simplifies to 2 3/4.

Let me know if this helps :)

5 0

3 years ago

Read 2 more answers

X - Y =4 - 2x+y=18 Linnear system

Dmitry_Shevchenko [17]

there is no answer to this so i can't help you.?

8 0

3 years ago

30km=?miles 44,000in=?cm 21ft=?m 0.75miles=?dam 133yards=mm Please show your work

kherson [118]

So there is this chart my teacher showed me to help find the answers to your questions I will send a picture but here are the answers

30km is 18.642m
44,000in is 111760cm
21ft is 6.401m
0.75m I’m not for sure sorry but
133yards is 121615.2mm

I hope that helps sorry I did not understand the one question 0.75m

5 0

3 years ago

Write a rule for the linear function in the table.

Ira Lisetskai [31]

Answer:

f(x) = 4 - x

Step-by-step explanation:

'f(x) = blah-blah' is the same as saying 'y = blah-blah', except it is also telling you that the x-y relationship is a function.

I notice that for each case, x + y = 4, so y = 4 - x

Or in 'function' terminology, f(x) = 4 - x

5 0

3 years ago

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ

steposvetlana [31]

Solution:

From Bernoulli equation

$\frac{1}{2}$ ρ $V_{1}^{2}$ + ρgh = $\frac{1}{2}$ ρ $V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$ are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$ $V_{1}$ = $\pi r_{2}^{2}$ $V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$

6 0

3 years ago