Answer:
<em>See the graph with both lines attached</em>
A) From the tables we see the constant change of values in the second column and the initial value is zero.
This represents a proportional relationship for both tables.
B) <u>Revel</u>
- Rate of change is 5 per hour, the equation is:
- y = 5x, where y is cost, x - is the number of hours
<u>Scooters</u>
- Rate of change is 6 per hour, the equation is:
- y = 6x, where y is cost, x - is the number of hours
C) <u>When x = 12</u>,
- y = 5*12 = $60 for Revel
- y = 6*12 = $72 for Scooters
D) <u>The point (4, 24) represents that:</u>
- The cost of 4 hours is $24
E) The second table has confusing titles (days and number of cell phones) but if we assume its same as the first table, then Revel has lower rate as 5 < 6.
Comparing the prices and the graph, Revel has better rate per hour.
Notice that 25+23+17+14+12+9=100, so among these students there are no 2 studying 2 subjects.
The probabilities of selecting students studying a certain subject are as follows
P(physics)=25/100
P(chemistry)=17/100
P(maths)=9/100
P(sociology)=23/100
P(political sciences)=14/100
P(anthropology)=12/100
since all the sets are disjoint, that is there are no common elements, and since all the students in consideration are enrolled in one these 6 subjects:
P(physics)+P(chemistry)+P(maths)+P(sociology)+P(political sciences)
+P(anthropology)=1
P(a)=P(sociology)+P(political sciences)+P(anthropology)+P(physics)
thus
P(a')=1-P(a)=P(chemistry)+P(maths)=17/100+9/100=26/100=0.26
Answer: 0.26
Answer:
34 grams
Step-by-step explanation:
If the remaining sample has 30.26 grams of radioactive substance, and 11% of it decayed, that means that 30.26 grams is 89% of the original. Let the original be x.
30.26=0.89x
Multiply both by one hundred
3026=89x
Divide both by 89
34=x
x=original, so the original was 34 grams.
<h3>
Determining the solution.</h3>
The solution(s) for the provided graph are (0, 6) and (6, 0)
<h3>What is a solution of a graph?</h3>
The solution of the <u>system of equations</u> is the <u>intersection</u> of the two equations.
<h3>Reasoning.</h3>
In the image provided, we can see two solutions (i<u>ntersection points)</u> that are intersected by a line and a circle.
<u>Those two solutions are:</u>