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2 years ago

What is the standard form of (5+i)(6-5i)/2i(-3+7i)

Mathematics

1 answer:

Margaret [11]2 years ago

6 0

Answer:

Since all parts of 2i(-3+7i) is in the denominator, you have to use an extra set of parentheses around the whole denominator. this is what you do

(5+i)(6−5i) / (2i(−3+7i))

= (30−25i+6i−5i^2) / (−6i+14i^2)

= (35−19i) / (−14−6i)

= (35−19i)(−7+3i) / (2(−7−3i)(−7+3i))

= (−245+105i+133i−57i^2) / (2(49−9i^2))

= (−188+238i) / (2(58))

= (−188+238i) / 116

= −47/29 + 119/58 iso that is how you do it

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Ons: Practice

Sedaia [141]

Answer:

$x = 18$

Step-by-step explanation:

Given

$\frac{x}{6} - 7 = -4$

Required

Solve for x

$\frac{x}{6} - 7 = -4$

Add 7 to both sides

$\frac{x}{6} - 7+7 = -4+7$

$\frac{x}{6} = -4+7$

$\frac{x}{6} = 3$

Multiply through by 6

$6 * \frac{x}{6} = 3 * 6$

$x = 3 * 6$

$x = 18$

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3 years ago

Find the perimeter pf a rectangular parking lot that has a width of 3x+2 and a length of 5x-7.*

uranmaximum [27]

Answer:

Perimeter is 16x - 10

Step-by-step explanation:

${ \sf{perimeter = 2(length + width)}} \\ p = 2 \{(5x - 7) + (3x + 2) \} \\ p = 2(8x - 5) \\ p = 16x - 10$

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3 years ago

P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball

Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .