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Vlad [161]
3 years ago
14

Write

Mathematics
2 answers:
8_murik_8 [283]3 years ago
6 0
hehehehehehehe.......

grin007 [14]3 years ago
3 0

8=2^3\\4=2^2\\\\8^3\div4^3\times2^5=(2^3)^3\div(2^2)^3\times2^5\\\\\text{Use}\ (a^n)^m=a^{n\cdot m}\\\\=2^{3\cdot3}\div2^{2\cdot3}\times2^5=2^9\div2^6\times2^5\\\\\text{Use}\ a^n\times a^m=a^{n+m}\ \text{and}\ a^n\div a^m=a^{n-m}\\\\=2^{9-6+5}=2^8

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3x + 7 equal 13x equal -2X equal to 2X equals 5
lianna [129]

3x + 7 = 13

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3 years ago
If the third term in an arithmetic sequence is 7 and the common difference is -5, what is the value of the fourth term?
kozerog [31]

Answer: 2

Step-by-step explanation:

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If P(C)=.29, P(D)=.47 and P(C and )= 0, are the events C and D mutually exclusive? Find P(C or D)
abruzzese [7]

Answer:

Step-by-step explanation:

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5 0
3 years ago
7.4 Practice
pickupchik [31]

The simulation of the medicine and the bowler hat are illustrations of probability

  • The probability that the medicine is effective on at least two is 0.767
  • The probability that the medicine is effective on none is 0
  • The probability that the bowler hits a headpin 4 out of 5 times is 0.3281

<h3>The probability that the medicine is effective on at least two</h3>

From the question,

  • Numbers 1 to 7 represents the medicine being effective
  • 0, 8 and 9 represents the medicine not being effective

From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two

So, the probability is:

p = 23/30

p = 0.767

Hence, the probability that the medicine is effective on at least two is 0.767

<h3>The probability that the medicine is effective on none</h3>

From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none

So, the probability is:

p = 0/30

p = 0

Hence, the probability that the medicine is effective on none is 0

<h3>The probability a bowler hits a headpin</h3>

The probability of hitting a headpin is:

p = 90%

The probability a bowler hits a headpin 4 out of 5 times is:

P(x) = nCx * p^x * (1 - p)^(n - x)

So, we have:

P(4) = 5C4 * (90%)^4 * (1 - 90%)^1

P(4) = 0.3281

Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281

Read more about probabilities at:

brainly.com/question/25870256

8 0
2 years ago
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