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4 years ago

The following equation f(x) = x³ + 0.5x² + 4.5 has only 1 real root. It lies between the interval [-1,0].

2 answers:

6 0

We are asked in this problem to find the real root of the polynomial f(x) = x3 + 0.5x2 + 4.5. in this case, to find the root of the equation, we equate the equation to zero.
<span>f(x) = x³ + 0.5x² + 4.5 = 0
2x3 + x^2 + 9 = 0
</span><span>x1=−1.83557
</span>x2=0.66779+1.41619∗i
x3=0.66779−1.41619∗i

answer is -1.84

Slav-nsk [51]4 years ago

4 0

Answer:b

Step-by-step explanation:

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Step2247 [10]

Step-by-step explanation:

The sequence is an AP with common difference -7.

Therefore the next 2 terms are 1 and -6.

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3 years ago

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Alex_Xolod [135]

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Step-by-step explanation:

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3 years ago

State University uses thousands of fluorescent light bulbs each year. The brand of bulb it currently uses has a mean life of 600

Ipatiy [6.2K]

Answer:

The test statistic is t = 2.5.

The p-value of the test is of 0.007 < 0.05, which means that the evidence supports the manufacturer's claim at the .05 significance level.

Step-by-step explanation:

Mean life of 600 hours. Test if it is more.

At the null hypothesis, we test if the mean is of 600 hours, that is:

$H_0: \mu = 600$

At the alternative hypothesis, we test if the mean is of more than 600 hours, that is:

$H_1: \mu > 600$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

600 is tested at the null hypothesis:

This means that $\mu = 600$

Suppose 100 bulbs were tested and found to have a mean of 625 hours with a standard deviation of 100.

This means that $n = 100, X = 625, s = 100$ .

Value of the test-statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{625 - 600}{\frac{100}{\sqrt{100}}}$

$t = 2.5$

The test statistic is t = 2.5.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 625 hours, which is a right-tailed test, with t = 2.5 and 100 - 1 = 99 degrees of freedom.

Using a t-distribution calculator, this p-value is of 0.007.

The p-value of the test is of 0.007 < 0.05, which means that the evidence supports the manufacturer's claim at the .05 significance level.

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3 years ago

What is the area of the polygon given below?

egoroff_w [7]

Answer:

186

Step-by-step explanation:

Cut it 6+7+11=24×5=120

6+11=66

120+66=186

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3 years ago

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Based on a random sample of 25 units of product X, the average weight is 102 lb and the sample standard deviation is 10 lb. We w

Dahasolnce [82]

Answer:

$t=\frac{102-100}{\frac{10}{\sqrt{25}}}=1$

$t_{crit}= 2.49$

Since the calculated value is lower than the critical value we have enough evidence to FAIL to reject the null hypothesis

So then we can conclude that the population mean is not significantly higher than 100

Step-by-step explanation:

Data given and notation

$\bar X=102$ represent the mean score for the sample

$s=10$ represent the standard deviation for the sample

$n=25$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha=0.01$ 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 determine if the mean is higher than 100, the system of hypothesis would be:

Null hypothesis: $\mu \leq 100$

Alternative hypothesis: $\mu > 100$

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".

Calculate the statistic

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

Calculate the critical value

We need to find the degrees of freedom first, given by:

$df= n-1= 25-1 = 24$

Since we are conducting a right tailed test we need to find a critical value on the t distribution with 24 degrees of freedom that accumulates 0.01 of the area on the right and we got:

$t_{crit}= 2.49$

Since the calculated value is lower than the critical value w have enough evidence to FAIL to reject the null hypothesis

So then we can conclude that the population mean is not significantly higher than 100

3 0

3 years ago