Answer:
a) P(X=2)= 0.29
b) P(X<2)= 0.59
c) P(X≤2)= 0.88
d) P(X>2)= 0.12
e) P(X=1 or X=4)= 0.24
f) P(1≤X≤4)= 0.59
Step-by-step explanation:
a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29
b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59
c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88
d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12
e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24
f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59
Answer:
negative, it'll be negative 2
1) To work out the result after a percentage is applied to a number, you work out 'the percentage number/100' and write the answer of this calculation down. 2) You THEN multiply this calculated number (called the 'percentage multiplier') by your orginal number (called the 'principal'). 3) The result of this calculation gives you the answer of what happens when your orginal number is multiplied by a certain percentage. 4) Thus, for a 2.99% increase work out: " (100+2.99)/100 "
The answer is 1.029 - SO THIS is your percentage multiplier. 5) Your earn 2.99% interest each year for 5 years. Thus you multiply your principal (10,500) by the calculated percentage multiplier (1.029), BUT you you multiply the principal by this multiplier 5 TIMES! 6) So, the calulation is:10,500 × (1.029^5). 7) This EQUALS $12,113.403198) Rounded to the nearest cent, the final amount in the account AFTER 5 YEARS = $12,113.40 8) THE ANSWER IS THUS: $12,113.40
Take the 4 out to get 4(x^2-12.5)
The velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.
Acceleration = 5 m/s2
Distance, s = 1 km = 1000m
Intital velocity, u=0 m/s
<h3>What is the first equation of motion?</h3>
We know that, Newton's equation of motion,
s = ut+ 1/2 at^2
1000 = 0 + 1/2 x 5 x t^2
1000 = 5/2 x t^2
2000/ 5 = t^2
t = 20 s
Also,a = (v-u)/t
5 = v-0/20
100 = v
v = 100 m/s
Hence, the velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.
Learn more about velocity here;
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