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Please help me with this. What is the slope of a line parallel to the line with equation 5x-2y=11?

kari74 [83]

The slope of a line parallel to another line must have the same slope as the other line. So therefore, we just want to first find the slope of this line.

To do this, we can switch it to slope intercept form, which is y = mx + b.

-2y = -5x + 11

Notice that the y doesn't have a variable next to it in slope intercept form. This means you have to divide all the sides to -2 to cancel out that -2.

-2y = -5x + 11
/-2 /-2 /-2

y = 5/2x + 11/-2

Since in the equation, y = mx + b, the number in the position of m = the slope, we know that 5/2 is the slope.

So therefore, the slope of a line parallel to the line with equation 5x - 2y = 11 must be 5/2.

6 0

2 years ago

Read 2 more answers

How many triangles exist with the given side lengths?12 in, 15 in, 18 in *

Mariulka [41]

Answer:

Only a scalene triangle would have the given side lengths making it a unique triangle.

Step-by-step explanation:

6 0

2 years ago

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lara31 [8.8K]

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: