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GaryK [48]
3 years ago
8

a vehicle travels 24km with a constant speed of 65 km/h and another 50 km with a constant speed of 80 km/h what was its average

speed
Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0
Average speed=totaldistance/totaltime

totaldistance=24+50=74km

total time
65/24=0.3692307 hours

80/50=1.6 hours
total time=0.3692307+1.6=1.9692307

averagespeed=74/1.9692307=37.5781263 km/h

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A random sample of 11 fields of rye has a mean yield of 20.1 bushels per acre and standard deviation of 7.66 bushels per acre. D
Lady bird [3.3K]

Answer:

17.1≤x≤23.1

Step-by-step explanation:

The formula for calculating the confidence interval is expressed as;

CI = x ± z*s/√n

x is the mean yield = 20.1

z is the 80% z-score = 1.282

s is the standard deviation = 7.66

n is the sample size =  11

Substitute

CI = 20.1 ± 1.282*7.66/√11

CI = 20.1 ± 1.282*7.66/3.3166

CI = 20.1 ± 1.282*2.3095

CI =  20.1 ±2.9609

CI = (20.1-2.9609, 20.1+2.9609)

CI = (17.139, 23.0609)

hence the required confidence interval to 1dp is 17.1≤x≤23.1

8 0
3 years ago
I don’t get it.. can anyone help?
tester [92]

Answer:

1. g x 5 = b

2. f x 12 = in

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4. input x 1.5 = output

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Step-by-step explanation:

7 0
3 years ago
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How do u solve this?​
kipiarov [429]

Answer:

20/9

Step-by-step explanation:

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What is 4 to the -2 power divided by 4 to the 4th power
igomit [66]

Answer:

4 to the -2 power divided by 4 to the 4th power

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<h3>(1/4096) is the right answer.</h3>
7 0
3 years ago
The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

Substituting into the ODE, we get

v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0
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Setting w=v', we end up with the linear ODE

w'x\ln x+(2+\ln x)w=0

Multiplying both sides by \ln x, we have

w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

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we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
w=\dfrac{C_1}{x(\ln x)^2}

Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
v=-\dfrac{C_1}{\ln x}+C_2

So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
4 0
3 years ago
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