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Ahat [919]
3 years ago
14

If f(x) = 2x2 - x, find f(-3).

Mathematics
1 answer:
forsale [732]3 years ago
7 0

f(x)=2x^2-x


Substitute the value of x = -3 to the equation:


f(-3)=2(-3)^2-(-3)=2(9)+3=18+3=21

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An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co
levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)  

2. 4.7048 \leq \sigma^2 \leq 16.1961

3. t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

t_{crit}=1.753

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

\chi^2 =24.9958

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

\bar X=26.31 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=16-1=15

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that t_{\alpha/2}=2.13

Now we have everything in order to replace into formula (1):

26.31-2.13\frac{2.8}{\sqrt{16}}=24.819    

26.31+2.13\frac{2.8}{\sqrt{16}}=27.801

So on this case the 95% confidence interval would be given by (24.8190;27.8010)  

Part 2

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=16-1=15

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

\chi^2_{\alpha/2}=24.996

\chi^2_{1- \alpha/2}=7.261

And replacing into the formula for the interval we got:

\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}

4.7048 \leq \sigma^2 \leq 16.1961

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:    

Null hypothesis:\mu \leq 25      

Alternative hypothesis:\mu > 25      

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)      

Calculate the statistic      

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

t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

Critical value  

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got t_{crit}=1.753

Conclusion      

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: \sigma \leq 2.3

H1: \sigma >2.3

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be \chi^2 =24.9958

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0
3 years ago
Kerrie uses the work shown below to determine 35% of 280.
lys-0071 [83]

For this case we must find 35% of 280.

Kerrie gets 10% and then 25%. So:

10% of 280:

280 ------------> 100%

x -----------------> 10%

Where "x" represents the amount given by 10% of 280.

x = \frac {10 * 280} {100}\\x = 28

Thus, 10% of 280 is 28.

Now, we find 25% of 280:

280 ------------> 100%

x -----------------> 25%

Where "x" represents the amount given by 25% of 280.

x = \frac {25 * 280} {100}\\x = 70

Thus, 25% of 280 is 70.

So we have that 35% of 280 is given by:

28 + 70 = 98

So, Kerrie's solution is correct.

Answer:

Option A

6 0
3 years ago
Read 2 more answers
Evaluate this. <br><br> SHOW FULL WORK!!!
faust18 [17]
Hi there!

2³ [ (15 - 7) × (4 ÷ 2) ] = 8 [ 8 × 2 ] = 8 × 16 = 128

Hence,
The required answer is 128

~ Hope it helps!
4 0
3 years ago
Read 2 more answers
What is the value of the discriminant for the quadratic equation -3=-x 2+2x
seraphim [82]
X=3,-1 your answer is right here
3 0
3 years ago
Kayla and latasha work for a department store the amount of sales they made over the last year was collected monthly in the data
tresset_1 [31]

Based on the information represented by the boxplot ;

  • Latasha's lowest sale amount = 50

  • Kayla's median is between 200 and 300

  • Latasha has a greater spread due to higher IQR value

1.) <em><u>The Lowest amount of sale made by Latasha in one month </u></em>

  • The minimum value is denoted by the starting position of the lower whisker on a boxplot.

  • Lowest amount of sale made by Latasha = 50

2.) <em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>made</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>Kayla</u></em><em><u> </u></em><em><u>:</u></em>

  • 50% of sales made marks the median value in a boxplot, it is denoted by the vertical line in between the box.

  • 50% of sales made by Kayla is between 200 and 300

  • With median sale value being 250

3.) <em><u>Spread</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>middle</u></em><em><u> </u></em><em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>:</u></em>

  • The measure of spread of the middle 50% of a distribution on a boxplot is the Interquartile range (IQR) of the distribution

  • IQR = Upper Quartile (Q3) - Lower quartile(Q1)

<u>For Latasha</u> :

  • Q3 = 450 (Endpoint of the box)
  • Q1 = 150 (starting point of the box)

  • IQR = 450 - 150 = 300

<u>For</u><u> </u><u>Kayla</u><u> </u><u>:</u><u> </u>

  • Q3 = 375 (Endpoint of the box)
  • Q1 = 100 (starting point of the box)
  • IQR = 375 - 100 = 275

  • Since, Latasha's IQR is greater than Kayla's, then Latasha has a greater mid 50% spread than Kayla.

Learn more :brainly.com/question/24582786

4 0
2 years ago
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