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lora16 [44]
3 years ago
15

I WILL GIVE BRAINLIEST!!! I REALLY NEED AN ANSWER ASAP!

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

Given:

The system of equations is

y=2x

y=-x+6

To find:

The solution of given system of equations.

Solution:

we have,

y=2x          ...(i)  

y=-x+6         ...(ii)

From (i) and (ii), we get

2x=-x+6

2x+x=6

3x=6

Divide both sides by 3.

x=2

Putting x=2 in (i), we get

y=2(2)

y=4

So, the solution is (2,4).

For y=2x,

x      y

0     0

1      2

For y=-x+6,

x       y

0     6

1     5

Plot these points on a coordinate plane and connect them by straight line as shown below.

From the below graph it is clear that both lines intersect each other at (2,4). So, the solution is (2,4).

Therefore, the solution of system of equations is (2,4).

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A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w
sveta [45]

Answer:

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

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".  

p_A represent the real population proportion for business  

\hat p_A =\frac{75}{400}=0.1875 represent the estimated proportion for Business

n_A=400 is the sample size required for Business

p_B represent the real population proportion for non Business

\hat p_B =\frac{137}{500}=0.274 represent the estimated proportion for non Business

n_B=500 is the sample size required for non Business

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

7 0
3 years ago
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