What is the Domain, Range, Decreasing, Increasing, Y-Intercept, X-Intercept, Maximum, and Minimum of the graph shown?

miskamm [114]

Answer:

the answer is D

Step-by-step explanation:

it is D because in normal y=mx+b form the given equasion is y= -1/2x + 3/2. for a line to be parallel to another line, they both must have the same slope, D is the only answer with the same slope as the slope from the given equasion

8 0

3 years ago

If y=1/2x+4, then find the equation of the line parallel through the point (6,3).

tatyana61 [14]

Answer:

x=2y

Step-by-step explanation:

random guess

5 0

3 years ago

If f(x) = x2 - 2x and g(x) = 6x + 4, for which value of x does (f+g)(x) = 0?

Law Incorporation [45]

Answer:

-2

Step-by-step explanation:

Let's plug your functions f(x)=x^2-2x and g(x)=6x+4 into (f+g)(x)=0 and then solve your equation for x.

So (f+g)(x) means f(x)+g(x).

So (f+g)(x)=x^2+4x+4

Now we are solving (f+g)(x)=0 which means we are solve x^2+4x+4=0.

x^2+4x+4 is actually a perfect square and is equal to (x+2)^2.

So our equation is equivalent to solving (x+2)^2=0.

(x+2)^2=0 when x+2=0.

Subtracting 2 on both sides gives us x=-2.

7 0

3 years ago

Read 2 more answers

A company estimates that its cost and revenue can be modeled by the functions C)-0.75x+20,000

lisabon 2012 [21]

Let C(x) = -0.75x + 20,000 and R(x)= -1.50x then the profit function exists noted as P(x) = R(x) - C(x)

P(x) = -1.50x - (-0.75)x + 20,000

P(x) = -0.75x + 20000

Therefore, the profit function exists -0.75x + 20000.

<h3>How to find profit function?</h3>

The profit function can be estimated by subtracting the cost function from the revenue function. Let profit be expressed as P(x), the revenue as R(x), the cost as C(x), and x as the number of items traded. Then the profit function exists noted as P(x) = R(x) - C(x).

Given:

C(x) = -0.75x+20,000 and R(x)= -1.50x

P(x) = R(x) - C(x)

= -1.50x - (-0.75)x + 20,000

= -1.50x + 0.75x + 20,000

Apply rule -(-a) = a

= -1.5x + 0.75x + 20000

Add similar elements:

-1.5 x + 0.75x = -0.75x

P(x) = -0.75x + 20000

Therefore, the profit function exists -0.75x + 20000.

To learn more about profit function refer to:

brainly.com/question/16866047

#SPJ9

8 0

2 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago