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3 years ago

what is te discontinuity of the function f(x) = the quantity of x squared plus 6x plus 8 all over x plus 4?

Mathematics

1 answer:

seropon [69]3 years ago

7 0

Answer:

A hole at x=-4.

Step-by-step explanation:

This is a fraction so we have to worry about division by zero.

The only time we will be dividing by 0 is when x+4 is 0.

Solving the equation

x+4=0 for x:

Subtract 4 on both sides:

x=-4

So there is either a vertical asymptote or a hole at x=-4.

These are the kinds of discontinuities we can have for a rational function.

If there is a hole at x=-4, then x=-4 will make the top zero and can be cancelled out after simplification.

If is is a vertical asymptote, x=-4 will make the top NOT zero.

Let's see what -4 for x in x^2+6x+8 gives us:

(-4)^2+6(-4)+8

16+-24+8

-8+8

Top and bottom are 0 when x=-4.

Let's see what happens after simplication.

We are going to factor a^2+bx+c if factorable by finding two numbers that multiply to be c and add up to be b.

So what 2 numbers together multiply to be 8 and add up to be 6.

I hoped you said 4 and 2 because (4)(2)=8 where 4+2=6.

$\frac{x^2+6x+8}{x+4}=\frac{(x+4)(x+2)}{x+4}=x+2$

We we able to cancel out that factor that was giving us x=-4 is a zero.

Therefore there is a hole at x=-4.

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The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs

MissTica

Answer:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

$s=8.43$ represent the sample deviation

$n=26$ sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z 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 is higher than 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu > 52$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

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

$p_v =P(t_{25}>1.8025)=0.0418$

Conclusion

7 0

3 years ago

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