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Anastaziya [24]
2 years ago
11

Use the given graph to determine the limit, if it exists.

Mathematics
1 answer:
Ahat [919]2 years ago
8 0

The limit in the given graph \lim_{x \to \ 2^{-} } f(x) is 3 and \lim_{x \to \ 2^{+} } f(x) is -2

Given graph of a function and we have to determine the limits when x tends to 2 minus and when x tends to 2 plus.

When we see the graph we can find that the graph is not of the linear function because it is not straight line.

From x=2 and onwards it gives values values of only -2 because it is parallel to x-axis at y=-2.From x=2 and leftwards it gives values values of only 3 because it is parallel to x-axis at y=3.

Hence the limit of the function whose graph is shown is 3 and -2.

Learn more about limits at brainly.com/question/27517662

#SPJ10

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A) If a + 5 = 15, then find the value of a​
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The answer is 10

Step-by-step explanation:

a + 5 =15

15-5=10

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Manchu has drawn a
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Height of each triangle: 14cm

Area of each triangle: 168cm²

Step-by-step explanation:

So we know that the base of the area is 24. Lets find the height.

336 ÷ 24 = 14

If he makes two triangles out of the rectangle that just means he cuts it in half.

To find the area of one triangle lets do:

336 ÷ 2 = 168

Let's double check our answer:

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Seems great!

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There are 45 children in a class. If
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20%

Step-by-step explanation:

Total = 45

Absentees = 9

Percentage  = (No. of absentees / total) × 100

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A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe
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Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  &#10;[tex]\alpha 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 to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case 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".  

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

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

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