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serg [7]
3 years ago
11

The maximum weight allowed per car on The Twister carnival ride is 263 pounds. Your friend weighs 86 pounds. To be able to ride

in a car together, how much can you weigh? Write and solve the inequality.
A. x + 177 (less than or equal to symbol) 263; at most 86 pounds
B. x - 86 (less than or equal to symbol) 263; at most 177 pounds
C. x + 86 (less than or equal to symbol) 263; at most 177 pounds
D. x + 86 (less than or equal to symbol) 177; at most 263 pounds

{the less than or equal to symbol is just < with a line under it} :) Pls help
Mathematics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

answer: C

Step-by-step explanation:

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919,827 to the nearest 1000, 100, and 100
OverLord2011 [107]

nearest 1000: 920,000

nearest 100: 919,800

nearest 10 (you said 100 but I think you mean 10): 919,830

8 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
A marker costs $0.99. A pad of paper costs $1.25 more than the marker, which is a more reasonable estimate for the total cost of
andrew11 [14]

Answer:

$3

Step-by-step explanation:

Because you have to also think about tax of the item. Plus add them both up and it costs $2.24 without tax. Which means it's already above $2.

8 0
3 years ago
Please help explanation need it
Zepler [3.9K]

Step-by-step explanation:

jejejebe. s shs sjs sibskkw

4 0
3 years ago
What is the sale price for your purchase if you have a 20% off coupon and you buy a shirt for $19.99 and pants for $10.
Andrew [12]
Answer: I believe it is $23.99

Explanation:
19.99+10= 29.99
29.99x0.2= 5.992
29.99-5.99= $23.99

I hope this helped!
4 0
3 years ago
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