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2 years ago

Round 33.825 to the nearest hundredths in place

Mathematics

2 answers:

zaharov [31]2 years ago

7 0

Answer:

33.83

Step-by-step explanation:

Nadusha1986 [10]2 years ago

4 0

Answer:

33.83

<h3>^Hope this helps, Hazel^</h3>

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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

Which of the following tables matches the graph below?

navik [9.2K]

Right answer is C

We can solve it with the formula

y=mx+b

b is the point where graphic starts

Table C shows us 0 by x, and -4 by y

So the graph starts at (0, -4 )

3 0

2 years ago

Can somebody explain how to do this question

evablogger [386]

Answer:

The answer is D

Step-by-step explanation:

Remember:

Contrapositive means that if you reverse the hypothesis and the conclusion the statement will mean the same thing

7 0

2 years ago

Please help

alexandr402 [8]

$\bf 3\sqrt[3]{2a} -6\sqrt[3]{2a}\qquad \stackrel{add ing~like-terms}{\implies }\qquad -3\sqrt[3]{2a}$

3 0

3 years ago

Read 2 more answers

f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans

Kruka [31]

Answer:

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

Step-by-step explanation:

We want to find the Riemann sum for $\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx$ with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)$

where $\Delta{x}=\frac{b-a}{n}$ .

Step 1: Find $\Delta{x}$

We have that $a=0, b=\frac{3\pi }{4}, n=6$

Therefore, $\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}$

Step 2: Divide the interval $\left[0,\frac{3 \pi}{4}\right]$ into n = 6 sub-intervals of length $\Delta{x}=\frac{\pi}{8}$

$a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b$

Step 3: Evaluate the function at the left endpoints

$f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3$

$f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386$

$f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964$

$f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527$

$f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0$

$f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527$

Step 4: Apply the Left Riemann Sum formula

$\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315$

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

5 0

2 years ago