Can somebody help me?

Write an equation in slope-intercept form of the line that passes through (-5, 19) and (5.13).

ruslelena [56]

Answer:

y=-3/5x+16

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(13-19)/(5-(-5))

m=-6/(5+5)

m=-6/10

simplify

m=-3/5

y-y1=m(x-x1)

y-19=-3/5(x-(-5))

y-19=-3/5(x+5)

y=-3/5x-15/5+19

y=-3/5x-3+19

y=-3/5x+16

3 0

2 years ago

What is the missing numerator?

fgiga [73]

4/5 - /7 =

4/5 - 2/7

5 * 7 = 35

7 * 4 = 28

5 * 2 = 10

28/35 - 10/35 = 18/35

5 0

2 years ago

Read 2 more answers

The distance between two cities on a map is 10 in. The actual distance between the cities is 135. What is the scale on the map?

spayn [35]

The scale on the map is 1 in = 13.5 units

135/10 = 13.5

4 0

3 years ago

Can someone please help me with this

vovangra [49]

Answer:

y = 5

z = 55

Step-by-step explanation:

90 - 35 = 55

z = 55

7y = 35

$\frac{7y}{7} = \frac{35}{7}$

y= 5

3 0

2 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

Since is a left tailed test the p value would be:

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

5 0

3 years ago