The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
#SPJ4
Answer: x = -7 or x = -4
Step-by-step explanation: When an equation has a squared term in it
such as x² + 11x + 28 = 0, it's called a polynomial equation.
The rule for solving a polynomial equation
is to first set it equal to zero, then factor.
Since this equation is already set equal to zero,
our first step is to factor the left side of the equation.
To factor x² + 11x + 28, we need the factors
of 28 that add to 11 which are 7 and 4.
So we have (x + 7)(x + 4) = 0.
Whenever two terms are multiplied together to equal zero,
this means that either one or the other must equal zero.
So if (x + 7)(x + 4) = 0, then either x + 7 = 0 or x + 4 = 0.
Solving each equation from here, we get x = -7 or x = -4.
So our solution is {-7, -4}.
Answer:
x = 118 units
Step-by-step explanation:
tan 50° = x/99
x = 99(tan 50°)
x = 117.9
Step-by-step explanation:
a. P(mites OR ticks) = P(mites) + P(ticks) − P(mites AND ticks)
P(mites OR ticks) = 0.80 + 0.30 − 0.20
P(mites OR ticks) = 0.90
b. P(no ticks | mites) = P(no ticks AND mites) / P(mites)
P(no ticks | mites) = (0.80 − 0.20) / 0.80
P(no ticks | mites) = 0.75
c. If two events are independent, then P(A and B) = P(A) × P(B).
P(mites AND ticks) = 0.20
P(mites) × P(ticks) = 0.80 × 0.30 = 0.24
They are not independent.
