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3 years ago

Given that the domain is all real numbers, what is the limit of the range for the function ƒ(x) = 42x - 100?

Mathematics

2 answers:

charle [14.2K]3 years ago

7 0

This function has a straight line graph and thus has no limit on the range.

Irina-Kira [14]3 years ago

4 0

Answer:

Step-by-step explanation:

The given function is

$f(x)=42x-100$

As you can notice, it's a linear function, and it's written in slope-intercept form. So, from this expression we can deduct that the slope is 42 and the y-intercept is at -100. We deduct this information, because of the form

$f(x)=mx+b$

Where $m$ is the slope and $b$ is the y-intercept, which in this case are

$m=42$ and $b=-100$ .

An important characteristic of linear function is that they domain and range sets are always all real numbers, the function is not restricted.

Therefore, the range of the given function is

R = (-∞,∞)

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Answer:

(x+4)(x-4)

Step-by-step explanation:

Here is a simple help for when you can take a square root of both terms:

(x+(squareroot)) x (x-(squareroot))

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4 x -4 = -16

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A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

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

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

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

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

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

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

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3 years ago