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nydimaria [60]
3 years ago
14

Three consecutive even numbers have a sum between 84 and 96. Write an inequality to find the three numbers. Let n represent the

smallest even number. Solve the inequality
Mathematics
2 answers:
yawa3891 [41]3 years ago
8 0

Answer:

28 30 and 32

Step-by-step explanation:

pantera1 [17]3 years ago
5 0
The numbers are: n, (n+2) and (n+4);
we can suggest this inequation:
84<n+(n+2)+(n+4)<96

Now, we solve this inequation:
84<n+(n+2)+(n+4)<96
84<3n+6<96

1)
3n+6>84
3n+6-6>84-6
3n>78
3n/3>78/3
n>26

2)
3n+6<96
3n+6-6<96-6
3n<90
3n/3<90/3
n<30

The solution is:
26<n<30

Then: n=28

The numbers will be: 28,30 and 32.

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a) P=0.558

b) P=0.021

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a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558

b) The approximate probability that at least 9 carry the gene is:

P(x\geq9)=1-P(x\leq 8)\\\\\\

P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

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Answer:

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Hope this helped! Please mark as brainliest! Thanks!!

8 0
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