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3 years ago

Examine the given sequence. Which statement is not correct?

Mathematics

2 answers:

Alja [10]3 years ago

4 0

Answer:

D) If c = 14.4, the relationship is exponential. a1 = 10 and an + 1 = an + 1.2 for n = {1, 2, 3, ...}.

Step-by-step explanation:

N76 [4]3 years ago

3 0

The answer I feel it is is choice C and the reason why is because

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What is the value of b2-4ac for the following equation? 5x2+7x=6

poizon [28]

Answer:

169

Step-by-step explanation:

Given data

The function is

5x^2+7x=6

rearrange

5x2+7x-6=0

From the above expression, we can see that

a=5

b=7

c=-6

Substitute into the expression b^2-4ac

=7^2-4*5*-6

= 49-(-120)

=49+120

=169

Hence the answer is 169

8 0

2 years ago

Help me please ill give brainlist

solmaris [256]

Answer:

106.8 in.

Step-by-step explanation:

The circumference of a circle:

C = 2πr

Substitute the given information into the formula.

C = 2(3.14)(17) = 106.8

5 0

3 years ago

Read 2 more answers

Complete the like terms to create an equvalent expression: -4r - 2r + 5

Marat540 [252]

-6r+5
the two like terms are the ones with R. combine -4r and -2r to get -6r

3 0

2 years ago

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

3 years ago

Confidence Interval Mistakes and Misunderstandingsâ€”Suppose that 500 randomly selected recent graduates of a university were as

kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

$n=500\\\bar x=8.6\\\sigma=2.2$

(a)

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

$8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)$

(b)

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

n = 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0

3 years ago