Answer:
11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they received a pneumococcal vaccination, or they did not. The probability of an adult receiving a pneumococcal vaccination is independent of other adults. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
70% of U.S. adults aged 65 and over have ever received a pneumococcal vaccination.
This means that 
20 adults
This means that 
Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination.
This is P(X = 12).
11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.
Answer:
We are given to find the length of side AB. We know that in a triangle, if two angles have equal measures, then the sides opposite to them are equal in length. Thus, the length of side AB is 6 units.
Hope this helps!!
PART1:
First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20
Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20
Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20
PART 2:
First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit
Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit
Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit
Hope this helps!!
Answer:
513 g of C₁₂H₂₂O₁₁ will contain 18 moles of Carbon atoms.
Step-by-step explanation:
English Translation
You would know how to say the number of moles of carbon atoms in 513 g of C₁₂H₂₂O₁₁ (Molar mass of C₁₂H₂₂O₁₁: 342 g.
Solution
From the molecular formula for the compound,
1 mole of C₁₂H₂₂O₁₁ consists of 12 moles of carbon atoms.
We just need to find the number of molesnof the mass of C₁₂H₂₂O₁₁ provided.
Number of moles = (mass)/(molar mass)
Mass of C₁₂H₂₂O₁₁ = 513 g
Molar mass of C₁₂H₂₂O₁₁ = 342 g
Number of moles of C₁₂H₂₂O₁₁ present
= (513/342) = 1.5 moles
1 mole of C₁₂H₂₂O₁₁ consists of 12 moles of carbon atoms.
1.5 moles of C₁₂H₂₂O₁₁ will contain (1.5×12/1) moles of carbon atoms, that is, 18 moles of Carbon atoms.
Hope this Helps!!!
0.8931/3 is equal to 0.2977
