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DaniilM [7]
3 years ago
12

I need the answers for 21 and 22

Mathematics
1 answer:
Trava [24]3 years ago
5 0

Answer:

21.b

22.c

Step-by-step explanation:

idk how to explain it lol I did mental math

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The answer to this question is C
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Given the prime factorisation of the each of
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The cube root of 287496 is 66
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Please help me simplify these expressions
Rina8888 [55]

Answer:

4. -2s^9

7. 8b^10

Step-by-step explanation:

First, focus on the number part of the expression. For 4), multiply -2 x -1 x -1 = -2. Then, look at the exponents. When you're multiplying exponents with the same base (in number 4, the base is s), you can add the exponents.

2+3+4 = 9; which makes it s^9. Multiply -2 x s^9, which gives you -2s^9.

Same thing for 7). Multiply all the numbers first. this gives you 8. (Remember two negatives equals a positive). Add all of the exponents of b together, which gives you 10. (when a variable has no exponent, it means that the exponent is 1.) Multiplying 8 and b^10 should give you 8b^10.

It may help to brush up on exponent rules and sign rules.

6 0
3 years ago
What is the inverse of f(x)=6x^2
likoan [24]

Answer:

f^-1 (x) = √ 6x/6 - √ 6x/6

Step-by-step explanation:

interchange the variables and solve for y.

sorry its a little hard to read i didnt have all the symbols on my keyboard

8 0
3 years ago
In a test of a printed circuit board using a random test pattern, an array of 14 bits is equally likely to be 0 or 1. Assume the
denis-greek [22]

Answer:

The probability that all 14 bits are 1s = 0.000061

Step-by-step explanation:

Probability that a bit is 1 is p = 0.5

Probability that a bit isn't 1 is q = 1 - 0.5 = 0.5

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 14

x = Number of successes required = 14

p = probability of success = 0.5

q = probability of failure = 0.5

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)¹⁴⁻¹⁴

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)⁰

P(X = 14) = 1 (0.000061)(1) = 0.000061

5 0
3 years ago
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