Answer: The answer is (b) a ≤ 3 + 2j; j ≥ 14, a ≤ 35.
Step-by-step explanation: Given that Anna is no more than 3 years older than 2 times Jamie’s age. Jamie is at least 14 and Anna is at most 35. We are to select the correct combination of inequalities among the given options.
Also, 'a' and 'j' are the possible ages of Anna and Jamie respectively. Therefore, according to the given information, we can write
Thus, the correct option is (b) a ≤ 3 + 2j; j ≥ 14, a ≤ 35.
Answer:
112
Step-by-step explanation:
The general form of all percentage questions is: the chunk= (some percentage) (of the whole), or c=p*w
We know that 18% of students are 8th grade and that is 126 students, so p = 0.18 and c=126 (126 students are 18% of the whole school)
126 = (0.18)w, divide both sides by 0.18
126/(0.18) = w = 700
9th graders are 16% of the school or 16% of 700 students
c = (0.16) 700 = 112 students
Answer:
D
Step-by-step explanation:
Labor productivity is is measured per hours.
Hence, the work he does [amount of claims] will be distributed amongst the amount of hours he works. The amount of dollar he uses <em>doesn't matter for labor productivity.</em>
<em />
Thus, his work is 6 claims. He does that is 8 hours. So claim/hr is:
6/8 = 0.75 claims per hour
Answer choice D is right.
IN finding the interest we need to use the following formula:
Interest= Principal x Rate x Time or I= PRT
Substitute values: I= [$20 + ($10 x $34)] -320 =$40
P= $320
R=?
T=10 months/year
I=PRT
Since R is a missing term, we will solve for R using this formula: R=I/PT
R= [<span>$20 + ($10 x $34)-320] / ($320 x 10 months)
T=10 months</span>÷12 months=0.83<span>
R= ($40)/ $320 X 0.83
R= 40/ 265.6
R=15.06024096</span>
<h2>Answer</h2>
0.43
<h2>Explanation</h2>
Remember that 
Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so 
. Now, the sum of all frequencies will be the sum of all the 10th graders, so 
. Let's replace the values:
And rounded to the nearest hundredth:
