What is 2 tens and 7 tenths as a decimal

Y = 1.10x + 5.00 <br> y = 1.50x + 3.00<br> (substitution)

nataly862011 [7]

Solving the system of equations $y = 1.10x + 5.00\\y = 1.50x + 3.00$ we get x=5 and y=10.5

Step-by-step explanation:

We need to solve the system of equations by substitutions.

The system of equations given are:

$y = 1.10x + 5.00\\y = 1.50x + 3.00$

Let:

$y = 1.10 x+5.00\,\,eq(1)\\y = 1.50 x+ 3.00\,\,eq(2)$

Putting value of y from eq(2) into eq(1)

$1.50 x+ 3.00=1.10 x+5.00\\1.50 x-1.10 x=+5.00-3.00\\0.4x=2.00\\x=2.00/0.4\\x=5$

So, value of x=5

Now putting value of x in eq 1 to find value of y:

$y = 1.10x + 5.00$

$y = 1.10(5) + 5.00$

$y = 5.5 + 5.00$

$y = 10.5$

So, value of y is 10.5

Solving the system of equations $y = 1.10x + 5.00\\y = 1.50x + 3.00$ we get x=5 and y=10.5

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

4 0

3 years ago

Read 2 more answers

What is 173 divided by 63?

Anni [7]

It is 2.74 and if you have to round it ,it is 3.

7 0

3 years ago

What is 1/4 - [-3/8] =

WINSTONCH [101]

$\frac{1}{4} - \frac{ - 3}{8} = \frac{5}{8}$
5/8 cross multiply on the top then multiply the denominator you get 20/32 reduce that it's 5/8

6 0

2 years ago

Read 2 more answers

Researchers are interested if a school breakfast program leads to taller children. Assume that the population of all 5 year-old

DedPeter [7]

Answer:

$39-1.96\frac{1}{\sqrt{25}}=38.608$

$39+1.96\frac{1}{\sqrt{25}}=39.392$

So on this case the 95% confidence interval would be given by (38.608;39.392)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=39$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1$ represent the population standard deviation

n=25 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$