Answer: A
Step-by-step explanation:
Your answer is A.
Value of 3.78 x 10=37.8
Value of 5.99 x 102=699.98
value of 7.35 x 10-12=61.5
Value of 9.46 x 10-10=84.6.
Least value to greatest value
3.78 x 10=37.8 Least value
7.35 x 10 - 12=61.5 Sencond least value
9.46 x 10 - 10=84.6 Sencond greatest value
5.99 x 102=699.98 Greatest value
Hope it helps have a great day:)
9514 1404 393
Answer:
21x² +20x +100 = 0
Step-by-step explanation:
We know the sum of the roots of x² +bx +c = 0 is -b, and their product is c. If the roots are α and β, then ...
The sum of the roots of the new equation will be ...
-b' = (α+1/β)+(β+1/α) = (α+β) +(1/α +1/β) = (α+β)(1 +1/(αβ))
The product of the roots of the new equation will be ...
c' = (α+1/β)(β+1/α) = αβ +2 +1/(αβ)
Using the above relations for (α+β) and αβ, we find that ...
-b' = (-b)(1 +1/c)
c' = c + 2 + 1/c
For the given equation, our definition of b and c is ...
b = 2/3
c = 7/3
so the new equation has values ...
b' = (2/3)(1 + 1/(7/3) = (2/3)(10/7) = 20/21
c' = 7/3 + 2 + 1/(7/3) = 13/3 + 3/7 = 100/21
So, the equation with the roots of interest is ...
x² +20/21x +100/21 = 0
Multiplying by 21 gives ...
21x² +20x +100 = 0
Answer:
(70-55)/4.5=y rounded down to the nearest whole number
Step-by-step explanation:
The first step is to take away the $55 from $70
70-55=15
Simba has $15 to spend on the yoga classes, to find the number of yoga classes she can attend, you divide 15 by the amount a yoga class costs, $4.50
15/4.50=3 1/3
Because she can not have 1/3 of a yoga class, she can attend 3 yoga classes in her budget.
Hope this helps, if you have any questions feel free to ask
Have a good day! :)
Answer:
0=0
1=10
2=15
3=20
Step-by-step explanation:
What I did was I multiplied each number by 5, then added 5.
Answer:
The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018
Step-by-step explanation:
The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula
In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.
The cummulative distribution function of W will be denoted by 
. The values of can be found in the attached file.
Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.
