Select all of the terms that are "like." xy2 2xy x2 -xy 2x2y 1/2xy

Mathematics

1 answer:

harina [27]3 years ago

6 0

Like terms will have the exact same variables and exponents...

like term are : 2xy , xy , 1/2xy

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Currency on hand = 100,000 Brazilian cruzeiros

Vladimir79 [104]

Answer: $18,552.9

Step-by-step explanation:

given data:

currency on hand = $100,000 Brazilian dollar.

curreach desired = us dollars.

if $1 USD = $5.39 Brazilian dollar.

therefore;

$100,000 Brazilia USD

= $100,000/5.39

= $18552.9 USD.

The desired amount in US dollars is $18,552.9

4 0

3 years ago

Read 2 more answers

Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

$f(x)=me^{-mx}\ ;\ m>0, x>0$

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
The mean of the distribution is: $\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20$ .
The standard deviation is: $\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20$

Compute the probability that a person is willing to commute more than 25 miles as follows:

$P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865$