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AleksandrR [38]
3 years ago
13

HURRY!

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
8 0

Answer:

Negative 15 x and 15 x

Step-by-step explanation:

i'm sorry if it is wrong please comment if it is so i can fix it

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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>

<em>13. What is the approximate value for the modal daily sales? </em>

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.

5 0
3 years ago
The finances of a group of pet owners were analyzed to determine how much they were spending on their pet(s) each year. A graph
ra1l [238]

Answer:

c.

Step-by-step explanation:

i did it

6 0
3 years ago
Aston challenges his fater to a 100-m race. Aston asks to race again but wants to be given a head start. How much of a start sho
photoshop1234 [79]
Ah hah ! You said in there "race AGAIN". Tell us something about how the first race turned out, and we may have some information we can use to answer with. Right now we got nuttn.
7 0
3 years ago
Can some on help me on this question?
mario62 [17]

Answer:

x=10 and x-7 is the answer

Step-by-step explanation:

a)2x-7=13

2x=13+7

2x=20

x=20/2

x=10

b)3x+4=25

3x=25-4

3x=21

x=21/3

x=7

i hope it will help you

6 0
3 years ago
Read 2 more answers
Last winter, the ratio of days with snow to day with no snow was 1.18. Write this ratio as a faction in simplest form
Alekssandra [29.7K]

to change any decimal to a fraction, we simply <u>use as many zeros on the denominator as there are decimals and lose the dot</u>.

1.18 has two decimals, so we'll use two zeros at the bottom, and lose the dot.


\bf 1.\underline{18}\implies \cfrac{118}{1\underline{00}}\implies \stackrel{simplified}{\cfrac{59}{50}}

6 0
3 years ago
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