Answer:
No. Only 1 to 100, logically.
Let's make a scenario to solve this.
So let's say I have an all yellow, 9 quarter spinner board and I spin. It will always land on yellow, so it has a 9/9 or 100% chance of landing on yellow.
Now we must create another scenario.
For instance, I have one bottle, its starting direction was North, and I spin it. It will have a 1/4 chance of landing on North, East, South, or West because it will land in a random direction. For the 4 diagonal directions <em>southeast, southwest, northeast, northwest</em> it will be the same because there are only 4 directions, therefore giving the probability a 1 in 4 chance of landing on either one. :)
So first write an inequality
360 + 90m > 810
-360 -360
90m > 450
÷90 ÷90
m > 5
Answer:
y has a finite solution for any value y_0 ≠ 0.
Step-by-step explanation:
Given the differential equation
y' + y³ = 0
We can rewrite this as
dy/dx + y³ = 0
Multiplying through by dx
dy + y³dx = 0
Divide through by y³, we have
dy/y³ + dx = 0
dy/y³ = -dx
Integrating both sides
-1/(2y²) = - x + c
Multiplying through by -1, we have
1/(2y²) = x + C (Where C = -c)
Applying the initial condition y(0) = y_0, put x = 0, and y = y_0
1/(2y_0²) = 0 + C
C = 1/(2y_0²)
So
1/(2y²) = x + 1/(2y_0²)
2y² = 1/[x + 1/(2y_0²)]
y² = 1/[2x + 1/(y_0²)]
y = 1/[2x + 1/(y_0²)]½
This is the required solution to the initial value problem.
The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.
For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.
With this, y has a finite solution for any value y_0 ≠ 0.
the answer is 4
just think of -4 as 4 in this question it is looking for the one closer and now that both of them are counting numbers 4 is less
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!
