Solve for y: 5x-3y<75

Find the measure of each angle indicated.

bekas [8.4K]

You have to decide all sides to figure out all the kissing pieces

7 0

3 years ago

A lawyer drives from her home, located 2 miles east and 3 miles north of the town courthouse, to her office, located 5 miles

LekaFEV [45]

Let the town courthouse be the origin of coordinate plane, the positive direction of x-axis is west and negative direction is east, the positive direction of y-axis is north and negative direction is south. Then a lawyer home has coordinates (-2,3) and lawyer office has coordinates (5, -21).

Use the distance formula $d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ to get the distance between lawyer's home and office:

 $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(-2-5)^2+(3-(-21))^2}=$
$=\sqrt{7^2+24^2} = \sqrt{49+576} = \sqrt{625} =25$ .
Answer: 25 miles.

7 0

4 years ago

Ignoring those who said they weren't sure, there were 297 men asked, and 183 said yes, they had driven a car when they probably

USPshnik [31]

Answer:

$z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998$

$p_v =2*P(z>3.998)=0.0000639$

With the most common significance levels used $\alpha= 0.1, 0.05, 0.01$ we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5

Step-by-step explanation:

Information given

n=297 represent the random sample of male taken

X=183 represent the men who said yes, they had driven a car when they probably had too much alcohol

$\hat p=\frac{183}{297}=0.616$ estimated proportion of men who said yes, they had driven a car when they probably had too much alcohol

$p_o=0.5$ is the value that we want to test

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Hypothesis to test

We need to conduct a hypothesis in order to test the claim that the majority of men in the population (that is, more than half) would say that they had driven a car when they probably had too much alcohol, and the system of hypothesis are:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

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

After replace we got:

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

$z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998$

Decision

We have a right tailed test so then the p value would be:

$p_v =2*P(z>3.998)=0.0000639$

With the most common significance levels used $\alpha= 0.1, 0.05, 0.01$ we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5

7 0

3 years ago

Stephen evaluated (6.34 times 10 Superscript negative 7 Baseline) (4.5 times 10 Superscript 3 Baseline). His work is shown below

Maksim231197 [3]

Answer:

The step 4 and 5 describe the errors Stephen made.

Step-by-step explanation:

Given : Stephen evaluated $(6.34\times 10^{-7})(4.5\times 10^3)$ . His work is shown below.