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3 years ago

12

Frank works in the supply room of a law firm. At the beginning of the month, he had 21 boxes of pens with 13 pens in each box. A

t the end of the month, there were 45 pens left. How many pens did he give out over the month?
A.
273
B.
79
C.
228
D.
318

2 answers:

Natali [406]3 years ago

5 0

It would be C.228 am I right or wrong?

Klio2033 [76]3 years ago

4 0

Answer: He gave out 228 pens

Step-by-step explanation:

At the beginning of the month, Frank had 21 boxes of pens with 13 pens in each box. This means that the total number pens that he Frank had at the beginning of the month is 13 × 21 = 273 pens.

At the end of the month, there were 45 pens left. The number of pens that he gave out will be the number of pens at the beginning of the month - the number of pens left at the end of the month. It becomes

273 - 45 = 228

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The low temperatures in Anchorage, Alaska, for a week were − 6, − 3, − 1, 1, − 3, − 9, and − 14 degrees Celsius. What two temper

Inessa05 [86]

The two temperatures could be removed so that the average of the remaining numbers is the same as the average of the original set of numbers are -1 and -9

Step-by-step explanation:

The formula of the average of a set of data is $A=\frac{S}{N}$ , where

S is the sum of the data in the set
N is the number of the data in the set

∵ The low temperatures in Anchorage, Alaska, for a week were

− 6, − 3, − 1, 1, − 3, − 9, and − 14 degrees Celsius

∴ The set of the data is { -6 , -3 , -1 , 1 , -3 , -9 , -14}

- Add the data in the set

∵ S = -6 + -3 + -1 + 1 + -3 + -9 + -14

∴ S = -35

∵ The set has 7 data

∴ N = 7

- Use the formula of the average above to find it

∴ $A=\frac{-35}{7}$

∴ A = -5

∴ The average of the original set of numbers is -5

To keep the average the same after removing two numbers multiply the average by the new number of data to find the new sum

∵ The number of data after removing two numbers is 5

∴ N = 5

∵ A = -5

- By using the rule of average

∴ $-5=\frac{S}{5}$

- Multiply both sides by 5

∴ -25 = S

∴ The new sum is -25

- Lets find which two numbers of have a sum of -10 because

the difference between -35 and -25 is -10

∵ -1 + -9 = -10

∵ -6 + -3 + 1 + -3 + -14 = -25

∴ We could removed -1 and -9

The two temperatures could be removed so that the average of the remaining numbers is the same as the average of the original set of numbers are -1 and -9

Learn more:

You can learn more about the average in brainly.com/question/10764770

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4 years ago

0.017 and 10, please retype the steps you use to set up the problem and type your steps to solve .

svetoff [14.1K]

Answer:

0.17

Step-by-step explanation:

0.017 times 10 will equal to 0.17 as you will need to move the decimal point forward by one

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3 years ago

In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortes days of the year va

gayaneshka [121]

Question:

In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortest days of the year vary by 2 hours 53 minutes from the equinox. In this year, the equinox falls on March 21. In this task, you'll use a trigonometric function to model the hours of daylight hours on certain days of the year in New York City.

- Find amplitude and the period of the function

- Create a trigonometric function that describes the hours of sunlight for each day of the year

- Then use the function you built to find how fewer daylight hours February 10 will have then March 21

Answer:

(a)

$A = 2.883$ --- Amplitude

$T = 365$ ---- Period

(b) Trigonometry function

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

(c) $Hours= 1.794$

Step-by-step explanation:

Given

$Average\ Sunlight = 12hr\ 8 min$

$Variance = 2hr\ 53min$

Solving (a): Amplitude (P) and Period (T)

The amplitude is the amount of time the longest and the shortest day vary.

So

$A = 2\ hr\ 53\ min$

Convert to hours

$A = 2\ + \frac{53}{60}$

$A = 2+0.883$

$A = 2.883$

The period (T) is the duration i.e 1 year

$T = 1\ year$

Assume no leap year

$T = 365$

Solving (b): Trigonometry function

The function follows a sinusoidal pattern and the general form is:

$f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))$

Where

$\mu = Average\ Value$

$\mu = 12\ hr 8\ min$

Convert to hours

$\mu = 12 + \frac{8}{60}$

$\mu = 12 + 0.133$

$\mu = 12.133$

$A = 2.883$ --- Amplitude

$T = 365$ ---- Period

$n = Equinox$

$n = March\ 21$

$March\ 21st = 80th\ day$

So:

$n= 80$

The function becomes:

$f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))$

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

Solving (c): Fewer daylight hours will Feb. 10 have.

$Feb\ 10 = 41st\ day$

So:

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

$f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[41 - 80])$

$f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[-39])$

$2\pi = 360^\circ$

So:

$f(41) = 12.133 + 2.883sin(\frac{360}{365}[-39])$

$f(41) = 12.133 + 2.883sin(-38.466)$

$f(41) = 12.133 - 2.883*0.6221$

$f(41) = 10.339$

The fewer daylight hours is the calculated as:

$Hours= Average - f(41)$

$Hours= \mu - f(41)$

$Hours= 12.133 - 10.339$

$Hours= 1.794$

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Answer:

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