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4 years ago

Before finding the surface area of a cylinder, you must convert all dimensions to the same unit of measure. True False

Mathematics

2 answers:

klio [65]4 years ago

7 0

Answer:

true

Step-by-step explanation:

Orlov [11]4 years ago

7 0

Answer:TRUE

Step-by-step explanation:

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A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

Allushta [10]

Answer:

4968.6 m from where it was fired
221.33 m/s

Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

(x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

To find the range, we can solve the equation y=0 for t, and use this value of t to find x.

Using the quadratic formula, we find t at the time of landing to be ...

t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

t ≈ 31.939 seconds

Then the horizontal travel is

x = 220·cos(45°)·31.939 ≈ 4968.6 . . . . meters

As it happens, the value under the radical in the above expression for time, when multiplied by s, is the vertical speed at landing. The horizontal speed remains s·cos(θ), so the resultant speed is the Pythagorean sum of these:

landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

6 0

3 years ago

The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a

nadezda [96]

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

$P(E|A)=0,\ P(E|B)=0.3,\ P(E|C)=0.4,\ P(E|D)=0.6$

First we calculate the probability that print quality is poor,

$P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)$

$P(E)=(0)(0.8)+(0.3)(0.02)+(0.4)(0.08)+(0.6)(0.1)$

$P(E)=0+0.006+0.032+0.06$

$P(E)=0.098$

a. Determine the probability of high ink viscosity given poor print quality.

$P(C|E)=\frac{P(E|C)P(C)}{P(E)}$

$P(C|E)=\frac{0.4\times 0.08}{0.098}$

$P(C|E)=\frac{0.032}{0.098}$

$P(C|E)=0.3265$

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

$P(A|E)=\frac{P(E|A)P(A)}{P(E)}$

$P(A|E)=\frac{0\times 0.8}{0.098}$

$P(A|E)=\frac{0}{0.098}$

$P(A|E)=0$

Probability of misaligned paper given poor quality is

$P(B|E)=\frac{P(E|B)P(B)}{P(E)}$

$P(B|E)=\frac{0.3\times 0.02}{0.098}$

$P(B|E)=\frac{0.006}{0.098}$

$P(B|E)=0.0612$

Probability of printer-head debris given poor quality is

$P(D|E)=\frac{P(E|D)P(D)}{P(E)}$

$P(D|E)=\frac{0.6\times 0.1}{0.098}$

$P(D|E)=\frac{0.06}{0.098}$

$P(D|E)=0.6122$

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

3 0

3 years ago

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If abc ~ def , what is the scale factor

Semmy [17]

Answer:

now dont

mnshdhhd

bdhxhxhdbhdjwuh

7 0

3 years ago

The force,

neonofarm [45]

$\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{array}{llll} \textit{"F" is inversely proportional}\\ \textit{to the square of "d"}\\ F=\cfrac{k}{d^2} \end{array}\qquad \textit{we also know that} \begin{cases} F=\stackrel{Newtons}{0.009}\\ d=\stackrel{meters}{2} \end{cases}$

$0.009=\cfrac{k}{(2)^2}\implies 0.009=\cfrac{k}{4}\implies 0.036=k~\hfill \boxed{F=\cfrac{0.036}{d^2}} \\\\\\ \textit{when F = 0.062, what is "d"?}~~~~~~0.062=\cfrac{0.036}{d^2} \\\\\\ d^2=\cfrac{0.036}{0.062}\implies d^2=\cfrac{18}{31}\implies d=\sqrt{\cfrac{18}{31}}\implies d\approx 0.76$

7 0

3 years ago

A family is trying to pick from different shades of paint for their living room. They have narrowed it down to 3 different blues

BARSIC [14]

The answer would be about a 35% chance of getting blue. I'm not 100% sure, but I do believe it is around this answer.

3 0

3 years ago

Read 2 more answers