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11Alexandr11 [23.1K]
3 years ago
5

Suzy regularly cuts across her neighbor's yard to get to school in the morning. How much distance does she save by cutting acros

s the lawn? Explain your reasoning.
The lengths are 12ft, 24 ft., and "Suzy's Path" would be the hypotenuse.
Mathematics
1 answer:
podryga [215]3 years ago
7 0
She would have to walk 12+24 which is 36 but she only has to walk:
√12²+24² = 26.8

36-26.8=9.2

She saves 9.2 ft

Hope this helps :)
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(02.06 MC)
Paul [167]
<span>The inequality |2x − 6| less than or equal to 10 can be expressed as two form like this to get two value of x
2x-6 </span>≤<span> 10
2x</span>≤10+6
x≤8<span>

-(</span>2x-6) ≤ 10<span>
2x-6 </span>≥<span> -10
2x</span>≥-10+6
x≥-4
<span>
The graph of the inequality would be: </span> 
x≤8 and x≥-4   
or   
-4≤x≤8 
5 0
4 years ago
Paul traded 13 baseball card to Dan for 4 new packs of 6 cards each. Paul lost 3 cards. He now has 75 cards. How many cards did
melomori [17]

Answer:

The number of cards Paul had at the beginning was 64 cards

Step-by-step explanation:

Here we have a word problem as follows

Paul traded 13 baseball card to Dan for 4 new packs of 6 card each

Number of cards received from Dan = 6 × 4 = 24

Number of cards traded to Dan = 13

Net number of cards traded = 24 - 13 = 11

Total number of cards Paul now has = 75 cards

Therefore since Paul gained 11 cards to make his total number of cards = 75, then;

The initial amount of cards Paul had = 75 - 11 = 64 cards.

3 0
3 years ago
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
Can someone help me on problem number 8?
kramer
List the coordinates,
Coordinates are shown as (x,y) 
If 2+2+7+7 is 16
You make those your side lengths 
Coordinates are (9,7) (2,7) (2,5) (9,5) 
they maybe slightly off because of the broken peice of paper
8 0
4 years ago
Hurry please!
muminat

Answer:

Johnny is correct because irrational numbers never end

3 0
3 years ago
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