Answer:
For this particular case they are interested on the amount of weight gained by randomly selecting some students, we need to remember that the weight can't be a discrete random variable since this random variable can take values on a specified interval and with decimals, so for this case the best conclusion is that we have a continuous data set.
Step-by-step explanation:
Previous concepts
We need to remember that continuous random variable mans that the values are specified over an interval in the domain, so is possible to have decimal values for the possible outcomes of the random variable.
By the other hand a discrete random variable only can take integers for the possible outcomes of the random variable over the specified domain.
Solution to the problem
For this particular case they are interested on the amount of weight gained by randomly selecting some students, we need to remember that the weight can't be a discrete random variable since this random variable can take values on a specified interval and with decimals, so for this case the best conclusion is that we have a continuous data set.
X^2 + 4x -12 = (x+6)(x-2)
so
(x+6)(x-2)
---------------- = x - 2
x + 6
f(x) = x -2
because
if x = -6
then y = -6 - 2 = -8
(-6,-8) is the order pair
Answer:
Need more information. What do you mean "Sixty people are invited to a party" ?
Step-by-step explanation:
Answer:
= 1, 2, 5, 8 and 26
Step-by-step explanation:
To find the first 4 terms, substitute n = 1, 2, 3, 4 into the formula
a₁ = 3(1) - 4 = 3 - 4 = - 1
a₂ = 3(2) - 4 = 6 - 4 = 2
a₃ = 3(3) - 4 = 9 - 4 = 5
a₄ = 3(4) - 4 = 12 - 4 = 8
The first 4 terms are
- 1, 2, 5, 8
Substitute n = 10
a₁₀ = 3(10) - 4 = 30 - 4 = 26
Answer:
Hypotenuse = 5√2 units
Step-by-step explanation:
Triangle ABC:
The legs are 5 units each
Opposite = 5unit
Adjacent = 5unit
Triangle ACB:
lengths of sides A C and C B are 5
The legs are 5 units each
Leg =AC = Opposite = 5unit
Leg = CB = Adjacent = 5unit
The hypotenuse for both ∆ACB is unknown. Since ∆ACB is a right angled triangle, we would apply Pythagoras theorem to find the hypotenuse.
Using Pythagoras theorem:
Hypotenuse ² = opposite ² + adjacent ²
Hypotenuse ² = leg² + leg²
AB = hypotenuse in ∆ACB
AB² = 5² + 5²
AB² = 25+25 = 50
AB = √50 = √(25×2)
AB = 5√2
Hypotenuse = 5√2 units
