282-p=27
I’m not sure if this is what you meant
I don't know what 'which of the following' is referring to, but the line has a positive slope, goes through the origin, and has a slope of 2/3.
Ok so this might be a bit confusing but I hope u understand it. Because the problem is in seconds, I tried to stick to seconds. In an hour there are 60 minutes and in those 60 minutes there are 3,600 seconds. Now in 25 seconds the mule deer can run 1/4 mile. So think for a second how many seconds it would take for it run a full mile. If u do the math then u would get 25*4 which is 100 seconds to run a full mile. Going back to our 3,600 seconds, all u have to do is divide it by 100 to get the miles. So 3,600/100 is 36. The mile deer runs 36 miles in an hour. I hope this was of help to u
Answer:
Step-by-step explanation:
Hello!
a)
The given information is displayed in a frequency table, since the variable of interest "height of a student" is a continuous quantitative variable the possible values of height are arranged in class intervals.
To calculate the mean for data organized in this type of table you have to use the following formula:
X[bar]= (∑x'fi)/n
Where
x' represents the class mark of each class interval and is calculated as (Upper bond + Lower bond)/2
fi represents the observed frequency for each class
n is the total of observations, you can calculate it as ∑fi
<u>Class marks:</u>
x₁'= (120+124)/2= 122
x₂'= (124+128)/2= 126
x₃'= (128+132)/2= 130
x₄'= (132+136)/2= 134
x₅'= (136+140)/2= 138
Note: all class marks are always within the bonds of its class interval, and their difference is equal to the amplitude of the intervals.
n= 7 + 8 + 13 + 9 + 3= 40
X[bar]= (∑x'fi)/n= [(x₁'*f₁)+(x₂'*f₂)+(x₃'*f₃)+(x₄'*f₄)+(x₅'*f₅)]/n) = [(122*7)+(126*8)+(130*13)+(134*9)+(138*3)]/40= 129.3
The estimated average height is 129.3cm
b)
This average value is estimated because it wasn't calculated using the exact data measured from the 40 students.
The measurements are arranged in class intervals, so you know, for example, that 7 of the students measured sized between 120 and 124 cm (and so on with the rest of the intervals), but you do not know what values those measurements and thus estimated a mean value within the interval to calculate the mean of the sample.
I hope this helps!
Answer:
5w + 4 = 99
Step-by-step explanation: