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The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other t

kap26 [50]

Answer:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If the z-score is higher than 2 or lower than -2, X is unusual.

In this question:

Mean = 40%. So $\mu = 0.4$

Standard error = 5%. So $\sigma = 0.05$

Is 46% unusual?

We have to find Z when X = 0.46. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.46 - 0.4}{0.05}$

$Z = 1.2$

1.2 is lower than 2, that is, it is only 1.2 standard deviations from the mean. So 46% is not unusual.

So the correct answer is:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

3 0

3 years ago

For the graph shown select the phrase that best represents the given system of equations

Drupady [299]

b) consistent and independent

8 0

4 years ago

Regan deposited $20 in a savings account earning 5% interest, compounded annually.

irga5000 [103]

$23
Mathias has $20 in a savings account. The interest rate is
5% per year and is not compounded. How much will he have in 3 years?

8 0

3 years ago

The volume of a sphere is decreasing at a constant rate of 116 cubic centimeters per second. At the instant when the volume of t

nirvana33 [79]

Answer:

$\frac{dr}{dt} = -1.325 \ cm/s$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Taking Derivatives with respect to time

Step-by-step explanation:

Step 1: Define

Given:

$V = \frac{4}{3} \pi r^3$

$\frac{dV}{dt} = -116 \ cm^3/s$

$V = 77 \ cm^3$

Step 2: Solve for r

Substitute: $77 = \frac{4}{3} \pi r^3$
Isolate r term: $\frac{77}{\frac{4}{3} \pi} = r^3$
Isolate r: $\sqrt[3]{\frac{77}{\frac{4}{3} \pi}} = r$
Evaluate: $2.63917 = r$
Rewrite: $r = 2.63917 \ cm$

Step 3: Differentiate

Differentiate the Volume Formula with respect to time t.

Define: $V = \frac{4}{3} \pi r^3$
Differentiate [Basic Power Rule]: $\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3 \cdot r^{3-1} \cdot \frac{dr}{dt}$
Simplify: $\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}$

Step 4: Find radius rate

Substitute in variables: $-116 \ cm^3/sec = 4 \pi (2.63917 \ cm)^2 \cdot \frac{dr}{dt}$
Isolate dr/dt rate: $\frac{-116 \ cm^3/s}{4 \pi (2.63917 \ cm)^2} = \frac{dr}{dt}$
Evaluate: $-1.3253 \ cm/s = \frac{dr}{dt}$
Rewrite: $\frac{dr}{dt} = -1.3253 \ cm/s$
Round: $\frac{dr}{dt} = -1.325 \ cm/s$