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hram777 [196]
2 years ago
9

Now, evaluate f (x) = 2x4 – 4x3 – 11x2 + 3x – 6 for x = –2. f (–2) =

Mathematics
2 answers:
Kipish [7]2 years ago
7 0

Answer:

8

Step-by-step explanation:

f(- 2) =  8

f(-2) = 2 (-2) ^ 4 - 4 (- 2) ^ 3 - 11 (- 2) ^ 2 + 3*(- 2) - 6 = 2 * 16 + 32 - 44 - 6 - 6 = 8

Therefore the value of f(x) is 8 at x = - 2

lorasvet [3.4K]2 years ago
3 0

Answer:

f(- 2) =  8

Step-by-step explanation:

To evaluate f(- 2) substitute x = - 2 into f(x) , that is

f(- 2) = 2(-2)^{4} - 4(- 2)³ - 11(- 2)² +3(- 2) - 6

       = 2(16) - 4(- 8) - 11(4) - 6 - 6

      = 32 + 32 - 44 - 12

      = 64 - 56

      = 8

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Answer:

If we are to make x small rugs, each of which takes 2 hours to dye, then the total time taken to dye the small rugs is 2x. Similarly for the y large rugs which each take 3 hours to dye, the total time for dyeing the large rugs is 3y. Therefore the total for all sizes of rugs is 2x + 3y. Finally, we have a maximum of 60 available hours for the dyeing, so the total time cannot exceed 60, and the final inequality is

2x + 3y < 60

Step-by-step explanation:

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2 years ago
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Answer:

l

Step-by-step explanation:

A sequence is an ordered list of numbers.....

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3 years ago
Please help me asap !!!
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Answer:

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Step-by-step explanation:

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2 years ago
Suppose that 500 parts are tested in manufacturing and 10 are rejected.
alexdok [17]

Answer:

a) z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation  

n=500 represent the random sample taken

X=10 represent the number of objects rejected

\hat p=\frac{10}{500}=0.02 estimated proportion of objects rejected

p_o=0.03 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:p=0.03  

Alternative hypothesis:p < 0.03  

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.

Calculate the statistic  

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

z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

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 one tailed left test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

3 0
3 years ago
[(25-11)+(15-9)] divided by 5 ;)
vovangra [49]
25-11=14
15-9=6
6+14=20
20/5
=4
8 0
3 years ago
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