Well, there are 52 cards in a deck, with 4 suits. there is a Jack and queen per suit, so that is a total of 8 jacks and queens in the deck. that probability looks like 8:52 as a ratio, or 2:13 simplified. so, you are likely to draw a Jack or queen 8 out of 52 times, or 2 out of 13 times
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
2%
take the absolute value of your (experimental-accepted then divide by the accepted) so (20-25)/25=.2
then multiply that number by 100 to get the percent, .2*100=2
Add 1
0.7+1=1.7
1.7+1= 2.7
etc., etc.
Answer:
a = $13.5
Step-by-step explanation:
Let a = adult tickets
Let c = children tickets
Translating the word problem into an algebraic equation;
<u>For the Martinez family;</u>
2a + 3c = $60
<u>For the Wright family;</u>
3a + 5c = $95.5
Thus, the simultaneous equations are;
..........equation 1
.........equation 2
We would use substitution method to solve;
From equation 2, we make a the subject of formula;
3a = 95.5 - 5c
a = (95.5 - 5c)/3
<em>Substituting the value of "a" into equation 1, we have;</em>
2[(95.5-5c)/3] + 3c = 60
Multiplying all through by 3;
2(95.5 - 5c) + 9c = 180
191 - 10c + 9c = 180
191 - c = 180
c = 191-180
c = $11
To find the value of a;
2a +3c = 60
<em>Substituting the value of "c" into the equation, we have;</em>
a = $13.5
<em>Therefore, the cost of an adult movie ticket is $13.5. </em>