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2 years ago

10

Working alone, Mr. Tough can grade the final exams in 12 hours. His assistant, Mrs. Nice, cam grade the same exams in 15 hours.

What fraction of the exams can Mr.Tough and Mrs.Nice grade in 1 hour if they work together?

1 answer:

s344n2d4d5 [400]2 years ago

6 0

Answer:

9n/60

Step-by-step explanation:

add 1/12 + 1/15 which = 5/60 + 4/60.

= 9/60

YAY!!!!!!!

You might be interested in

Jumbo eggs in Australia, on average, are supposed to weigh 68g. Margot is in charge of a quality control test that involves weig

Andrej [43]

Answer:

0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean weight of 67g and a sample standard deviation of 1g.

This means that $\mu = 67, \sigma = 1$

Sample of 4

This means that $n = 4, s = \frac{1}{\sqrt{4}} = 0.5$

What is the probability that the mean weight of 4 eggs in a package is less than 68.5g?

This is the pvalue of Z when X = 68.5. So

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

By the Central Limit Theorem

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

$Z = \frac{68.5 - 67}{0.5}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g

6 0

3 years ago

Read 2 more answers

The population of a certain species of fish has a relative growth rate of 1.1% per year. It is estimated that the population in

Juliette [100K]

Answer:

(a) n(t) = P(0)*e^(0.010939940t)

(b) 12,674,681 (nearest unit)

(c) 14 years (nearest year)

Step-by-step explanation:

rate = 1.1% / year = 1.011

(a)

P(0) = 12,000,000 = population in 2010

In compound interest format, after t years

P(t) = P(0)* (1.011)^t

Given format = P(0)* e^(rt)

therefore

e^(rt) = 1.011^t use law of exponents

(e^r)^t = 1.011^t

e^r = 1.011

r = log_e(1.011) = 0.010939940 (to 9 decimal places)

required formula is

n(t) = P(0)*e^(0.010939940t)

(b)

in 2015,

P(0)=12000000, n = 5 (years after 2010)

n(5) = 12000000*e^( 0.010939940 * 5 ) = 12,674,680.6 = 12,674,681 (nearest unit)

(c)

to reach 14 million, we equate

n(t) = 14,000,000

12,000,000 *e^(0.010939940*t) = 14,000,000

e^(0.010939940*t) = 14000000/12000000 = 7/6

take log on both sides

0.010939940*t = log(7/6)

t = log(7/6) / 0.010939940 = 14.091 years = 14 years to the nearest year.

See graph attached. Y-axis is in millions, x-axis is in years.

4 0

4 years ago

maksim [4K]

1. you would take 36×1/4 and then it would equal 9

2. you would take 24×1/4 and then it would equal 6

3 0

3 years ago

Dmitrij [34]

Answer:

aeafscgefgdderhgdthdg

8 0

3 years ago

Write the expression using rational exponents. Then simplify and convert back to radical notation.

ioda

Answer:

The radical notation is $3x\sqrt[3]{y^2z}$

Step-by-step explanation:

Given

$\sqrt[3]{27 x^{3} y^{2} z}$

Step 1 of 1

Write the expression using rational exponents.

$\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}$

$=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}$

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

4 0

3 years ago