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3 years ago

USE SUBSTITUTION METHOD TO FIND X AND Y PLEASE HELP ME QUICKLY

Mathematics

1 answer:

valina [46]3 years ago

7 0

I’m not sure if you need the steps but x is -4 and y is -1

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3 years ago

Use the theorem on inverse functions to prove that f and g are inverse functions of each other. f(x) = 3x − 2 g(x) = x + 23

nignag [31]

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Step-by-step explanation:

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3 years ago

A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting

jenyasd209 [6]

Answer:

$P(Z> 3.13) = 0.000874$

Step-by-step explanation:

A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and obtaining an average greater than 3.6.

We can denote the population mean with the symbol $\mu$

According to the information given, the data have a population mean:

$\mu = 3.2$ .

The standard deviation of the data is:

$\sigma = 0.7$ .

Then, from the data, a sample of size $n = 30$ is taken.

We want to obtain the probability that the sample mean is greater than 3.6

If we call

$\mu_m$ to the sample mean then, we seek to find:

$P(\mu_m> 3.6)$

To find this probability we find the Z statistic.

$Z = \frac{\mu_m-\mu}{\sigma_{\mu_m}}$

Where:

Where $\sigma_{\mu_m}$ is the standard deviation of the sample

$\sigma_{\mu_m} = \frac{\sigma}{\sqrt{n}}$

$\sigma_{\mu_m} =\frac{0.7}{\sqrt{30}}\\\\\sigma_{\mu_m} = 0.1278$

$P(\frac{\mu_m-\mu}{\sigma_{\mu_m}}> \frac{3.6-3.2}{0.1278})$

Then:

$Z = \frac{3.6-3.2}{0.1278}\\\\Z = 3.13$

The probability sought is: $P(Z> 3.13)$

When looking in the standard normal probability tables for right tail we obtain:

$P(Z> 3.13) = 0.000874$

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3 years ago

Please this is urgent!!<br>Solve for x:<br>a(x-b)+bx+c=a+b

Arada [10]

X= a+b+c+bx/a(x-b) but I’m not completely sure

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3 years ago

(1 points)

Iteru [2.4K]

Pls help ASAP i don’t have time!!!!! Pls help ASAP i don’t have time!!!!!

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3 years ago