A linear equation has the x multiplied by a number, while exponential function has the x as a power
The force created by the object is 3.75 Newton.
Given
Mass of the object (m) = 1250 grams
Convert 1250 grams to kilograms by dividing it by 1000.
Mass (m) = 1.25 kg
Acceleration of the body (a) = 3 
From Newton's second law we know 
By Substituting in the above equation we get
Force = 1.25 x 3
Force = 3.75 
It can also be written as force = 3.75 N
Hence Force created by the body is 3.75 Newton.
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Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The equation of the given line is
7x+4y=3
4y = - 7x + 3
y = -7x/4 + 3/4
Comparing with the slope intercept form, slope = -7/4
If the line passing through the given point is perpendicular to the given line, it means that its slope is the negative reciprocal of the slope of the given line.
Therefore, the slope of the line passing through (6,-9) is 4/7
To determine the intercept, we would substitute m = 4/7, x = 6 and y = -9 into y = mx + c. It becomes
- 9 = 4/7×6 + c = 24/7 + c
c = - 9 - 24/7 = -87/7
The equation becomes
y = 4x/7 - 87/7
Answer:
80
Step-by-step explanation:
Angle VSU = 100. Angles TSU is a linear pair with angle VSU so adding them together equals 180. 180-100=80.
Answer:
E(X) = 6
Var(X) = 3.394
Step-by-step explanation:
Let X represent the number of carp caught out of the 20 fishes caught. Now, if we are to assume that each
of the (100, 20) ways to catch the 20 fishes will be equally likely.
Thus, we can say that X fulfills a hypergeometric
distribution with parameters as follows;
n = 20, N = 100, k = 30
Formula for expected mean value in hypergeometric distribution is;
E(X) = nk/N
E(X) = (20 × 30)/100
E(X) = 6
Formula for variance is;
Var(X) = (nk/N) × [((n - 1)(k - 1)/(N-1))) + (1 - nk/N)]
Var(X) = ((20 × 30)/100) × [((20 - 1)(30 - 1)/(100 - 1)) + (1 - (20 × 30/100)]
Var(X) = 6 × 0.5657
Var(X) = 3.394
