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3 years ago

The Great Wall is more than 20,000 kilometers long about 3/4 of it is considered properly preserved

1 answer:

7 0

97. The great wall is more than 20 000 kilometers long. And about ¾ of it is considered as properly preserved. I guess you are asking about the value of ¾ which is properly preserved in the said length of the great wall. => ¾ = 0.75 Thus, the formula would be like this: => 20 000 * .75 = 15 000 Thus, the ¾ value of great wall that is preserved is equals to 15 000 kilometers long,

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Help , please and thanks

ira [324]

Answer:

891,962,172,409

try that

Step-by-step explanation:

5 0

3 years ago

Plz Help! 30 Points!

Ira Lisetskai [31]

45.2 is the constant hope this helps! please give branliest

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3 years ago

Read 2 more answers

A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum

Ivan

Answer:

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solution

$y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solving the differential equation

$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

Initial conditions:

$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

$y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48$

The rumor reaches 80% at 8.48 days.

8 0

3 years ago

A new gas- and electric-powered hybrid car has recently hit the market. The distance traveled on 1 gallon of fuel is normally di

Doss [256]

Answer:

a) P [ Z > 70 ] = 0.1075 or 10.75 %

b) P [ Z < 60 ] = 0.1038 or 10.38 %

c) P [ 55 ≤ Z ≤ 70 ] = 0.8882 or 88.82 %

Step-by-step explanation:

Normal Distribution μ = 65 and σ = 4

a) The probability of the car travels more than 70 miles per gallon is:

P [ Z > 70 ] = (Z-μ) ÷ σ ⇒ P [ Z > 70 ] = (70-65) ÷4 P [ Z > 70 ] = 1.25

the point 1.25 corresponds at values, from left tail up to 70 so we must go and look for the area for the point 1.24 which is 0.8925. Then we have the area or probability of all cars traveling up to 70 miles therefore we have to subtract 1 -0,8925

P [ Z > 70 ] = 0.1075 or 10.75 %

b)The probability of the car travels less than 60 miles per gallon is:

P [ Z < 60 ] = ( 60 - μ ) ÷ σ ⇒ P [ Z < 60 ] = (60-65)÷ 4 P [ Z < 60 ] = -1.25

Again -1.25 corresponds to 60 miles per gallon threfore we move to the left and find for point -1.26 which area is 0.1038 so

P [ Z < 60 ] = 0.1038 10.38 %

c) P [ 55 ≤ Z ≤ 70]

For point Z = 70 or 1.25 (case a above) P [ Z ≤ 70] = (70-65)÷4

P [ Z ≤ 70] = 1.25

In this case we got the whole area from the left tail up to 1.25

P [ Z ≤ 70] = 0.8944 (includes the area of the point from the left tail up to the point assocciated to 55 miles and for that reason we have to subtract that area)

P [ Z ≥ 55 ] = (55-65) ÷ 4 P [ Z ≥ 55 ] = -2.5 and the area is 0.0062

So P [ 55 ≤ Z ≤ 70 ] = 0.8944 - 0.0062 = 0.8882

8 0

3 years ago

Will give brainliest!

fiasKO [112]

Let's give some distances
100 up 100 flat 100 down meters
then return 100 up 100 flat 100 down

time required (distance/rate = time ) 200/100 + 200/120 + 200 / 150 = 5 hr

distance 600

distance / time = average speed = 600/5 = 120 m/s

5 0

3 years ago