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3 years ago

How would you simplify (3+i)(4+2i)?

Mathematics

2 answers:

const2013 [10]3 years ago

7 0

Answer:

Step-by-step explanation:

Use FOIL: (a + b)(c + d) = ac + ad + bc + bd

and i² = -1

$(3+i)(4+2i)\\\\=(3)(4)+(3)(2i)+(i)(4)+(i)(2i)\\\\=12+6i+4i+2i^2\qquad\text{combine like terms}\\\\=(12-2)+(6i+4i)\\\\=10+10i$

Lunna [17]3 years ago

7 0

Answer:

11 + 10i

Step-by-step explanation:

Note that we're multiplying two binomials here, and thus may use the FOIL method. This requires four multiplications, as follows:

3*4 + 3*2i + 4i - 1 (The last term is 1 because i(i) = i^2 = -1.)

Simplifying, we get:

12 + 6i + 4i -1, or:

11 + 10i

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A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. I

Whitepunk [10]

Answer:

a) P(X≥6) = 0.9614

b) P(7≤ X ≤9) = 0.628

c) P(X≤5) = 0.0386

Step-by-step explanation:

This can be solved using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where p = probability of success

q = probability of failure = 1-p

n = number of trials

x = number of successful trials

We have p = 70% = 0.7

n = 12

a) Find the probability that the number who disapprove of smoking pot daily is no less than 6 i.e. P(X≥6). This can be calculated as:

P(X≥6) = 1 - P(X<6)

= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)]

= 1 - [¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷]

= 1 - (0.000000531 + 0.0000148 + 0.0001909 + 0.00148 + 0.00779 + 0.02911)

= 1 - 0.0386

P(X≥6) = 0.9614

b) P(7≤ X ≤9) = P(X=7) + P(X=8) + P(X=9)

= ¹²C₇ (0.7)⁷(0.3)¹²⁻⁷ + ¹²C₈ (0.7)⁸(0.3)¹²⁻⁸ + ¹²C₉ (0.7)⁹(0.3)¹²⁻⁹

= 0.158 + 0.231 + 0.239

P(7≤ X ≤9) = 0.628

c) P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

= ¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷

P(X≤5) = 0.0386

4 0

3 years ago

8.The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb. The

Sladkaya [172]

Answer:

The value is c = 21.1445.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb.

This means that $\mu = 12, \sigma = 3.5$