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2 years ago

What is the product of (2i) and (5+3i)?

Mathematics

1 answer:

Evgesh-ka [11]2 years ago

5 0

The value of the product expression given as product of (2i) and (5+3i) is 10i -6

<h3>How to determine the product?</h3>

The product expression is given as:

product of (2i) and (5+3i)

Rewrite the expression as:

(2i) * (5+3i)

Open the brackets

(2i) * (5+3i) = 10i + 6i^2

Evaluate

(2i) * (5+3i) = 10i + 6(-1)

This gives

(2i) * (5+3i) = 10i -6

Hence, the value of the product expression given as product of (2i) and (5+3i) is 10i -6

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Is the gfc of a pair of numbers ever equal to one of the numbers

cricket20 [7]

Answer:

yes, they can be at times

Step-by-step explanation:

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3 years ago

Find coordinates of the focus, the length of the latus rectum, and the coordinates of its endpoint for y^2= 4x

d1i1m1o1n [39]

Answer:

a) Focus is (1,0)

b) Length of Latus rectum is 4

c) The endpoints of the latus rectum is (1,-2) and (1,2)

Step-by-step explanation:

The given parabola has equation

${y}^{2} = 4x$

When we compare to

${y}^{2} = 4px$

We have

$4px = 4x$

This implies,

$p = 1$

The focus of this parabola is at (p,0).

Therefore the focus is (1,0)

b) The length of the latus rectum is given by

$|4p|$

From a) part we found p to be 1.

Substitute p=1 to obtain:

$|4 \times 1| = 4$

c) The focus is the midpoint of the latus rectum.

Since the focus is (1,0), we substitute x=1 into the equation to see intersection of the latus rectum and the parabola.

${y}^{2} = 4(1)$

This means that:

${y}^{2} = 4$

Take square root:

$y = \pm \sqrt{4}$

$y = \pm2$

$y = - 2 \: or \: y = 2$

Therefore the end of latus rectum are at (1,-2) and (1,2)

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3 years ago

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of

s344n2d4d5 [400]

Answer:

There is a 91.15% probability that their mean life will be longer than 12 years.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

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.

In this problem, we have that:

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.1 years. This means that $\mu = 13.2, \sigma = 3.1$ .