First term = (2)5^(-2*1) =2/25
2nd term = (2)5^(-2*2) = 2/ 625
3rd term = 2(5)^(-2*3) = 2/ 15625
If we divide the 3rd term by the 2nd term or the 2nd term by the first we get the common ratio
Common ratio = 1/25 or 0.04 as a decimal fraction.
Answer:
8 cookies were left after the bake sale. 72 were sold.
Step-by-step explanation:
64*1.25= 80
80*.9= 72
8
125% of 64 is equal to 80.
90% of 80 is equal to 72.
The number left was 10% of the number he made. The number he made was 125% of 64, so the number left was 8
10% × 125% × 64 = 0.125 × 64 = 8
thats all i could really come up with at the top of my head
<u>Answer:</u> The probability that Xavier and Yvonne can solve a problem but Zelda cannot is 
<u>Step-by-step explanation:</u>
We are given:
Probability of success of Xavier, 
Probability of failure of Xavier, 
Probability of success of Yvonne, 
Probability of failure of Yvonne, 
Probability of success of Zelda, 
Probability of failure of Zelda, 
We need to calculate:
The probability that Xavier and Yvonne can solve the problem but Zelda cannot, we use:

Hence, the probability that Xavier and Yvonne can solve a problem but Zelda cannot is 
Answer:
3/6 or 1/2
Step-by-step explanation:
3 copies of 1/6
of = ×
3 × 1/6
3/6
to the least
1/2
Answer:
x = 14
y = 6
Step-by-step explanation:
<em>*</em><em>Property</em><em>*</em>
In a parallelogram:
Opposite sides are equal.
Since, the given figure is a parallelogram we can apply this property and equate:
- 1) 2y + 1 = 13
- 2) x = y + 8
<u>1) 2y + 1 = 13</u>
<em>subtracting</em><em> </em><em>1</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>sides</em><em>:</em>
=> 2y + 1 - 1 = 13 - 1
=> 2y = 12
<em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>2</em><em>:</em>
=> 2y/ 2 = 12/ 2
<em>On</em><em> </em><em>the</em><em> </em><em>LHS</em><em>,</em><em> </em><em>2</em><em> </em><em>gets</em><em> </em><em>canceled</em><em> </em><em>because</em><em> </em><em>it's</em><em> </em><em>available</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>numerator</em><em> </em><em>as</em><em> </em><em>well</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>denominator</em><em> </em>
<em>On</em><em> </em><em>the</em><em> </em><em>RHS</em><em>,</em><em> </em><em>2</em><em> </em><em>gets</em><em> </em><em>canceled</em><em> </em><em>when</em><em> </em><em>12</em><em> </em><em>is</em><em> </em><em>factorised</em><em> </em><em>to</em><em> </em><em>6</em><em> </em><em>×</em><em> </em><em>2</em><em>:</em>
<h3>=> y = 6</h3>
<u>2</u><u>)</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>y</u><u> </u><u>+</u><u> </u><u>8</u>
We've got the value of y from equation 1).
<em>Substituting it here:</em>
=> x = 6 + 8
<h3>=> x = 14 </h3>
Answer:
Hence, we got the values of x and y as:
