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3 years ago

How can I skip more helppppppppppppppppppppppp

Engineering

2 answers:

julia-pushkina [17]3 years ago

5 0

Answer:

answer someone else's question and you will never have to skip again

Explanation:

AfilCa [17]3 years ago

4 0

Answer: skip what

Explanation:

You might be interested in

9 b. A sign (including the post and base) weighs 40 pounds and is

tigry1 [53]

Answer:

The weight should be added to the base of the sign to keep it from tipping is 65.6 lb

Explanation:

Given data:

A sigh weighs 40 pounds

Suported by an 18 in x 18 in square

Force of the wind 13.2 lb

Questions: Will the sign tip over, if yes, how much evelnly distributed weight should be added to the base of the sign to keep it from tipping, W = ?

The sign and the post have a length of 6 ft. You need to calculate the distance from the edge to the middle point:

18/2 = 9 in = 0.75 ft

Force acting in the base (40 lb):

$F=\frac{40*0.75}{6} =5lb$

The weight should be added to the base:

$(40+W)*\frac{0.75}{6} =13.2\\W=65.6lb$

8 0

4 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

An automotive Battery is rated at120 A-h. This means that under certain test conditions, it canoutput 1 A at 12 V for 120 hours.

Hitman42 [59]

Answer:

Part A:

In W-h:

Energy Stored=1440 W-h

In Joules:

$Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ$

Part B:

In W-h:

Energy left=240 W-h

In Joules:

Energy left= 8.64*10^5 J

Explanation:

Part A:

We are given rating 120A-h and voltage 12 V

Energy Stored= Rating*Voltage (Gives us units W-h)

Energy Stored=120A-h*12V

Energy Stored=1440 W-h

Converting it into joules (watt=joules/sec)

Energy Stored= $1440 Joules * \frac{3600sec}{h}h$

Energy Stored=5184000 Joules

$Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ$

Part B:

Energy used by lights for 8h=150*8

Energy used by lights for 8h=1200W-h

Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h

Energy left=1440-1200

Energy left=240 W-h

Energy left= $240 Joules * \frac{3600sec}{h}h$

Energy left=864000 Joules

Energy left= 8.64*10^5 J

3 0

4 years ago

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD = 30 mg/L × dilution factor

Original BOD = 30 mg/L × 10 = 300 mg/L

$L_o = \frac{BOD}{1-e^{-5t}}$

here $L_o$ is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day

$L_o = \frac{300}{1-e^{-5(0.20)}}$

$L_o = 474.59 \ mg/L$

3 0

4 years ago

Airbags will deploy in a head on collision but not in a collision that occurs from angle

Aneli [31]

Answer:

Airbags will deploy in almost any angle.

Explanation:

Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.

8 0

3 years ago