Ask question

Ask question

All categories

3 years ago

9

In estimating the mean speed of a roadway segment,a pilot sample of 100 vehicles at this location yield a sample standard deviat

ion of 4.8 miles/ hr

1 answer:

Oxana [17]3 years ago

8 0

Answer:

See Explanation

Explanation:

Given

$n = 100$

$\sigma= 4.8$

Required

The question is incomplete.

Given the sample size and the standard deviation, a likely question could be to calculate the standard error (SE) of the mean

This is calculated as:

$SE = \frac{\sigma}{\sqrt n}$

Substitute values

$SE = \frac{4.8}{\sqrt{100}}$

Take the positive square root of 100

$SE = \frac{4.8}{10}$

$SE = 0.48$

You might be interested in

Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

or

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

or

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

or

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

or

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

or

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

or

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

or

$K_v=12.34$

3 0

4 years ago

A 65% efficient turbine receives 2 m^3/s of water from a reservoir. The reservoir water surface is 45 m above the centerline of

laiz [17]

Answer:

$P_{out} = 508.071 kW$

Given:

efficiency of the turbine, $\eta$ = 65% = 0.65

available gross head, $H_{G}$ = 45 m

head loss, $H_{loss}$ = 5 m

Discharge, Q = $2 m^{3}$

Solution:

The nozzle is 100% (say)

Available power at the inlet of the turbine, $P_{inlet}$ is given by:

$P_{inlet} = \rho Qg(H_{G} - H_{loss})$ (1)

where

$\rho$ = density of water = 997 $kg/m^{3}$

acceleration due to gravity, g = $9.8 m^{2}$

Using eqn (1):

$P_{inlet} = 997\times 2\times 9.8(45 - 5) = 781.65 kW$

Also, efficency, $\eta$ is given by:

$\eta = \frac{P_{out}}{P_{inlet}}$

$0.65 = \frac{P_{out}}{781.648\times 1000}$

$P_{out} = 0.65\times 781.648\times 1000 = 508071 W = 508.071 kW$

$P_{out} = 508.071 kW$

3 0

3 years ago

Your uncle has given you a newmonitor for your computer. When you attempt to connect it, you notice that none of the ports on th

juin [17]

Answer:

Go on amazon and look for HDMI -> (VGA/DVI/HDMI) what ever you need adapters.

6 0

3 years ago

Ihjpr2 ywjegnak'evsinawhe2'qwmasnh ngl,;snhy

WITCHER [35]

Answer:

ummm ok?

Explanation:

6 0

3 years ago

Read 2 more answers

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th

Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

8 0

3 years ago