x1=0.35*(x1)+0.37*(x2)+0.19*(x3) x2=0.01*(x1)+0.11*(x2)+0.15*(x3)+0.05*(x4) x3=0.01*(x2)+0.1*(x3)+0.02*(x4) x4=0.64*(x1)+0.51*(x2)+0.56*(x3)+0.93*(x4) x1+x2+x3+x4=1 solving equations

0.0822 0.0781 0.0741 0.0704 0.0668 0.0601 0.054012 0.0746 0.0705 0.0666 0.0628 0.0593 0.0527 0.046813 0.0681 0.0640 0.0601 0.0565 0.0530 0.0465 0.040814 0.0626 0.0585 0.0547 0.0510 0.0476 0.0413 0.035715 0.0578 0.0538 0.0499 0.0463 0.0430 0.0368 0.031516 0.0537 0.0496 0.0458 0.0423 0.0390 0.0330 0.027817 0.0500 0.0460 0.0422 0.0387 0.0354 0.0296 0.024718 0.0467 0.0427 0.0390 0.0355 0.0324 0.0267 0.021919 0.0438 0.0398 0.0361 0.0327 0.0296 0.0241 0.019520 0.0412 0.0372 0.0336 0.0302 0.0272 0.0219 0.017524 0.0329 0.0290 0.0256 0.0225 0.0197 0.0150 0.011328 0.0270 0.0233 0.0200 0.0171 0.0146 0.0105 0.007532 0.0226 0.0190 0.0159 0.0133 0.0110 0.0075 0.005036 0.0192 0.0158 0.0129 0.0104 0.0084 0.0053 0.003340 0.0166 0.0133 0.0105 0.0083 0.0065 0.0039 0

Performing the second approximation, ,x3 = x2 – [( x1 – x2) / (f(x1) – f(x2))]f(x2)=(- 0.234375) – [(1 – 0.25)/(-3 – (- 0.234375))](- 0.234375)= 0.186441Hence, f(x3) = 0.074276

. If 6P i=1 xi = 3 and 6P i=1 x2 i = 10 Evaluate a. 6P i=1 (3xi + 2) b. 6P i=1 (xi + 3)2 a. Evaluate 4P i=1 (2xi + 3) b. If 4P i=0 xi = 2 and nP i=4 x2 0 = −3 Evaluate a. 4P i=0 (3xi − 1)2 b. 4P i=0 (xi − 2)2 c. 4P i=0 x2 i + 2x1 + 3
1

0.16(2x - 1) = 0.8  *1 pointx=2x=3x=4x=5

Based on the below table, answer the following question:xP(x)30.0540.4050.3060.1570.10 Find P(X>3)Question 3Answera.0.95b.0.15c.0.3