Write 3.25 as a divisiob problem represented by a fraction

A die is rolled. Write each answer as a fraction, decimal, and percent. Show your work.

dalvyx [7]

A. 1/6, .16666666, 16.6666666%
B. 4/6, .64, 64%
C. 5/6, .80, 80%

3 0

4 years ago

A chocolate bar is produced with an advertised mass of 200 g.

Elan Coil [88]

Answer:

The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.

This means that $\mu = 200.4, \sigma = 0.05$

a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.

pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 - 0.05 = 199.9.

X = 200.9

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

$Z = \frac{200.9 - 200.4}{0.05}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 199.9

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

$Z = \frac{199.9 - 200.4}{0.05}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

8 0

3 years ago

A student’s course grade is determined by averaging 4 exams. The student’s grades on the first 3 exams are 85, 87, and 89. What

Nady [450]

Answer:

a). At least 99 marks should be scored by the student to get an A.

b). Student has to score at least 59 and less than 99 grades in the 4th exam to get a B.

Step-by-step explanation:

Student's grades on the first three exams are 85, 87, and 89.

Since student's course grade is to be determined by averaging 4 exams.

So average of 4 exams = $\frac{85+87+89+x}{4}$

Here x is the grade obtained by the student in 4th exam.

If the average of 4 exams to get an A is at least 90 so the expression will be

$\frac{85+87+89+x}{4}\geq 90$

$\frac{261+x}{4}\geq 90$

261 + x ≥ 360

x ≥ 360 - 261

x ≥ 99

At least 99 marks should be scored by the student to get an A.

To get a B (Average of at least 80) the expression will be

$80\leq \frac{85+87+89+x}{4}< 90$

$320\leq(261+x)< 360$

320-261 ≤ x < 360-261

59 ≤ x < 99

Therefore, student has to score at least 59 and less than 99 grades in the 4th exam to get a B.

8 0

4 years ago

The function g(x)= 112 Ln (0.121x) + 2011 models the year in which the population of New York City will equal x million people.

stellarik [79]

We are given function: $g(x)= 112\ ln (0.121x) + 2011$ .

Given function models a particular year of population of New York City.

x represents population of New York City ( In millions).

We need to estimate the population of New York City in 2020.

Because g(x) function represents a particular year of population of New York City and we are given year 2020, so we need to replace g(x) by 2020 and solve for x.

Replacing g(x) by 2020, we get

$2020= 112\ ln (0.121x) + 2011$ : <em>This is the required equation to estimate the population of New York City in 2020</em>