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pochemuha
3 years ago
10

Tyler’s brother earns $12 per hour. The store offers him a raise— a 5% increase per hour. After the raise, how much will Tyler’s

brother make per hour?
Mathematics
2 answers:
Artyom0805 [142]3 years ago
6 0
Tyler's bother will now make $12.60 per hour
rjkz [21]3 years ago
5 0

Answer: $12.60

Step-by-step explanation:

We are given that , Tyler’s brother earns $12 per hour.

Also, The store offers him a raise— a 5% increase per hour.

it means there is an 5% increase in his earnings.

Now, Increase in earnings per hour = 5% of  $12

= 0.05\times\$12= \$0.60

⇒Increase in earnings per hour = $0.60

Now, After the raise, Tyler’s brother will make per hour= Previous earnings + Increase in earning

= $12 +  $0.6= $12.60

Hence, After the raise, Tyler’s brother will make $12.60 per hour.

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What is the approximate value for the modal daily sales?
Aleksandr [31]

Answer:

Step-by-step explanation:

Hello!

<em>The table shows the daily sales (in $1000) of shopping mall for some randomly selected  days </em>

<em>Sales 1.1-1.5 1.6-2.0 2.1-2.5 2.6-3.0 3.1-3.5 3.6-4.0 4.1-4.5 </em>

<em>Days 18 27 31 40 56 55 23 </em>

<em>Use it to answer questions 13 and 14. </em>

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To determine the Mode of a data set arranged in a frequency table you have to identify the modal interval first, this is, the class interval in which the Mode is included. Remember, the Mode is the value with most observed frequency, so logically, the modal interval will be the one that has more absolute frequency. (in this example it will be the sales values that were observed for most days)

The modal interval is [3.1-3.5]

Now using the following formula you can calculate the Mode:

Md= Li + c[\frac{(f_{max}-f_{prev})}{(f_{max}-f_{prev})(f_{max}-f_{post})} ]

Li= Lower limit of the modal interval.

c= amplitude of modal interval.

fmax: absolute frequency of modal interval.

fprev: absolute frequency of the previous interval to the modal interval.

fpost: absolute frequency of the posterior interval to the modal interval.

Md= 3,100 + 400[\frac{(56-40)}{(56-40)+(56-55)} ]= 3,476.47

<em>A. $3,129.41 B. $2,629.41 C. $3,079.41 D. $3,123.53 </em>

Of all options the closest one to the estimated mode is A.

<em>14. The approximate median daily sales is … </em>

To calculate the median you have to identify its position first:

For even samples: PosMe= n/2= 250/2= 125

Now, by looking at the cumulative absolute frequencies of the intervals you identify which one contains the observation 125.

F(1)= 18

F(2)= 18+27= 45

F(3)= 45 + 31= 76

F(4)= 76 + 40= 116

F(5)= 116 + 56= 172 ⇒ The 125th observation is in the fifth interval [3.1-3.5]

Me= Li + c[\frac{PosMe-F_{i-1}}{f_i} ]

Li: Lower limit of the median interval.

c: Amplitude of the interval

PosMe: position of the median

F(i-1)= accumulated absolute frequency until the previous interval

fi= simple absolute frequency of the median interval.

Me= 3,100+400[\frac{125-116}{56} ]= 3164.29

<em>A. $3,130.36 B. $2,680.36 C. $3,180.36 D. $2,664</em>

Of all options the closest one to the estimated mode is C.

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Step-by-step explanation:

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