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3 years ago

Find the determinant of the following matrix. A =-2 63 5

Mathematics

1 answer:

Blababa [14]3 years ago

6 0

Answer:

$|A| = - 28$

Step-by-step explanation:

Given

$A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]$

Required

Determine the determinant

For a two by two matrix, A such that:

$A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

The determinant |A| is:

$|A| = a * d - b * c$

So, in

$A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]$

The determinant is:

$|A| = -2 * 5 - 6 * 3$

$|A| = -10 - 18$

$|A| = - 28$

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Martin deposits $700 into a savings account that gains interest at a rate of 5.5% annually. If he waits 7 years to withdraw the

Travka [436]

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Use the multiplier 1.055

final = original x multiplier^n

where 'n' is the number of years

Sub the values in

final = 700 x 1.055^7

final = 1018.27

Subtract the values:

1018.27 - 700 = 318.27

Thus, he earns more than $275

He earns $43.27 more

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4 0

3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

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 particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$