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2 years ago

If you know that R' (20)=50 and R''(20)=-4, approximate R'(24)

Mathematics

1 answer:

vladimir1956 [14]2 years ago

4 0

Approximate R' by using a linear approximation: for x close enough to 24, you have

R'(x) ≈ L(x) = R' (20) + R'' (20) (x - 20)

Then

R' (24) ≈ R' (20) + R'' (20) (24 - 20) = -4 + 50 (24 - 20) = 196

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When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se

ryzh [129]

Answer:

a) Joint ptobability distribution

$\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}$

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

$P(x,y)=P_x(x)*P_y(y)$

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

$\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}$

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

$P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56$

We can calculate P(X<= 1) * P(Y<=1)

$P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56$

Both calculations give the same result.

c) Probability of no violations

$P(X+Y=0)=P(0,0)=0.3$

d) P(X + Y <= 1)

$P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53$

5 0

3 years ago

Algebra question, please show work :))

WITCHER [35]

I dont know why but a moderator deleted my answer but I hope that you got it. It was actually A. since it was the quotient of 9 not the quotient of x.

Have a great day hon and I hope that this helped you out a lil:)

5 0

3 years ago

Read 2 more answers

Use differentials to estimate the amount of paint needed (in m3) to apply a coat of paint 0.04 cm thick to a hemispherical dome

pogonyaev

The amount of paint is the volume of paint needed.

The amount of paint needed is 0.065312 cubic meters

<h3>How to determine the amount of paint needed</h3>

The volume of a hemisphere is:

$V = \frac 23\pi r^3$

Differentiate the above equation

$V' = 2\pi r^2 r'$

The above equation represents the estimate of the amount of paint needed.

Where:

r represents the radius (r = 52/2 m)
r' represents the thickness (r' = 0.04 cm)

So, we have:

$V' = 2\pi r^2 r'$

$V' = 2 * 3.14 * (52/2\ m)^2 * (0.04cm)$

$V' = 2 * 3.14 * (26\ m)^2 * (0.04cm)$

Express cm as m

$V' = 2 * 3.14 * (26\ m)^2 * (0.0004m)$

$V' = 0.065312 m^3$

Hence, the amount of paint needed is 0.065312 cubic meters

If you know that​ R' (20)=50 and​ R''(20)=-4, approximate​ R'(24)

If you know that R' (20)=50 and R''(20)=-4, approximate R'(24)