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3 years ago

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For this question, you will be provided with data related to the count of website sessions by day for the past one hundred days.

You are now asked to create a forecast for the next sixty days using this data. The data for this problem is included in as a .csv file in the attached packet.

1 answer:

svet-max [94.6K]3 years ago

6 0

Answer: Idk

Explanation:

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The assembly consists of two blocks A and B, which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must d

Sedaia [141]

Answer:

5.70 m

Explanation:

3sₐ + sB = L

3∆sₐ = - ∆sB

3vₐ = -vB

3*3 = -vB

vB = -9 ms⁻²

T₁ + V₁ = T₂ + V₂

(0 + 0) + (0 + 0) = 1/2(20)(3)² + 1/2(30)(-9)² + 20*9.81*(sB/3) + 30*9.81* sB

0 = 90 + 1215 - 228.9sB

-1305 = - 228.9sB

sB = 1305/228.9 = 5.70 m

6 0

3 years ago

Read 2 more answers

The speed of sound in air is proportional to the square root of the absolute temperature. If the speed of sound is 349 m/s when

REY [17]

Given:

Let the speed of sound be represented by 'v' then

v ∝ $\sqrt{T}$ (1)

$v_{1}$ = 349 m/s

$v_{2}$ = 340 m/s

$T_{1}$ = 20°C = 273+20 = 293 K

Formulae used:

1) °C = K + 273

2) K = °C - 273

3) °F = 1.8°C + 32

4) °R = °F + 459.67

Solution:

From eqn (1),

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

$T_{2}$ = $T_{2} = (\frac{v_{2}}{v_{1}})^{2}T_{1}$

$T_{2} = (\frac{340}{349})^{2}{293}$ = 278.08 K

Now, Usinf formula (1), (2), (3) and (4) respectively, we get

1) T = 293 K

2) T = 293 -278.8 = 5.08°C

3) T = 1.8(5.08) + 32=41.14°F

4) T = 41.14 + 459.67 = 500.81°R

5 0

3 years ago

How does sea navigation work?

ahrayia [7]

Answer:

a clock

Explanation:

you use a clock in water

3 0

3 years ago

A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

Alex Ar [27]

Answer:

Throat diameter $d_2$ =28.60 mm

Explanation:

Bore diameter $d_1=63mm$ ⇒ $A_1=3.09\times 10^{-3} m^2$

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004 $\frac{m^3}{s}$

Coefficient of discharge $C_d$ =0.8

We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$h=x(\dfrac{S_m}{S_w}-1)$

$S_m$ =13.6 for Hg and $S_w$ =1 for water.

$h=0.235(\dfrac{13.6}{1}-1)$

h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

⇒ $d_2$ =28.60 mm

So throat diameter $d_2$ =28.60 mm

4 0

3 years ago

Section lines represent surfaces exposed by a cutting plane<br> a. True b. False

oksian1 [2.3K]

Answer:

these is very interesting question but I still know the answer is a. True

7 0

3 years ago