Solve for x and show work

Which of the following is an example of a function?

Leokris [45]

A. Not a function since we have the points (-1,0) and (-1,3). The x value repeats while the y values are different. Plugging in x = -1 leads to multiple outputs, which shows we don't have a function. To get a better visual, you can plot the points. You'll see that the points form a vertical line. Therefore, this graph fails the vertical line test.

B. This is a function. All of the x values are different. The points, when graphed, pass the vertical line test. In other words, its impossible to draw a vertical line through any two points on the graph. Any x input here leads to exactly one y output.

C. This is not a function. Like choice A, we have repeated x values. In this case we have (4,6) and (4,7). The x value x = 4 repeats itself leading to multiple outputs y = 6 and y = 7

D. This is also not a function. The x value x = 0 repeats itself. The graph fails the vertical line test.

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Answer: Choice B

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4 0

3 years ago

the area of rectangular park is 3/5 square mile, the length of the park is 7/8 mile,what is the width of the park

Luba_88 [7]

Area=LengthxWidth so 3/5=7/8xWidth. To get width by itself you must divide both sides by 7/8. When dividing fractions you can flip the fraction and multiply so multiply (3/5)(8/7). To do this just multiply the numerators and denominators to get 24/35 so width = 24/35

8 0

3 years ago

Which value of y will make the inequality y < -1 false ?

scoray [572]

Hello!

Since y has to be less than -1 to make it false y has to be greater than -1

The answer 2 or just any number that is greater than -1

Hope this helps!

6 0

2 years ago

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

weqwewe [10]

Answer:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

Step-by-step explanation:

1) Data given and notation

n=180 represent the random sample taken

X=125 represent the number of americans between 17 to 24 that not qualify for the military

$\hat p=\frac{125}{180}=0.694$ estimated proportion of americans between 17 to 24 that not qualify for the military

$p_o=0.75$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ 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 less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :

Null hypothesis: $p\geq 0.75$

Alternative hypothesis: $p < 0.75$

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:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

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

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

$p_v =P(z$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

6 0

2 years ago

The tax on a property with an assessed value of $90,000 is $1,200. What is the assessed value of a property if the tax is $2,200

Ivanshal [37]

I'm not 100% sure but this is how I would try.
If you don't get a better answer, check this w/your teacher.

x/100 *90,000 = 1200

90,000x = 120,000

x = 1.3% tax percentage

1.3/100x = 2200

1.3x = 220,000

x = $169,230.77 assessed value

5 0

3 years ago