Step-by-step explanation:
GIVEN: Two digit number means we will have to choose from numbers 0-9 for the second digit and 1-9 for the first digit. (since zero can't be in the tens place)
Formula: We will use m x n. That is, m is the number of elements we can use for the first digit and n will be the number of elements we can use for the second digit.
Sample Space: To get the probability, we must find the sample space or all the outcomes.
m - 9 elements (1-9)
m - 9 elements (1-9)n - 10 elements (0-9)
m x n = 9x10 =90
90 is our sample space.
A. An odd number
Conditions: For an odd number, the second number will determine whether or not it is an odd number. For the first digit, we can use 1-9.
Elements:
m - 9 elements (1-9)
m - 9 elements (1-9)n - 5 elements (1,3,5,7,9)
m x n=9x5=45
SOLVE (probability)
Probability= possible outcomes/sample space
P=45/90
P=1/2
B. Larger than 75.
Conditions: The first digit could be equal to 7 or greater. But n will only be greater than 5 if and only if m=7 and n could be any number if m>7.
Elements 1:
m - 1 element (7)
m - 1 element (7)n - 4 elements (6,7,8,9)
m - 1 element (7)n - 4 elements (6,7,8,9)m x n=1x4=4
Elements 2:
m - 2 elements (8,9)
m - 2 elements (8,9)n - 10 elements (0,1,2,3,4,5,6,7,8,9,)
m - 2 elements (8,9)n - 10 elements (0,1,2,3,4,5,6,7,8,9,)mxn=2x10=20
TOTAL: 4+20=24
SOLVE (PROBABILITY)
Probability= possible outcomes/sample space
P=24/90
P=4/15
C. multiple of 5
Condition: the second digit must be either 0 or 5 and the first digit could be any digit.
Elements:
m - 9 elements (1-9)
m - 9 elements (1-9)n - 2 elements (0,5)
m x n =9x2=18
SOLVE (PROBABILITY)
P=18/90
P=1/5
D. An even number smaller than 40
Condition: The first digit could be equal to 4 but the second digit must be even (excluding zero). Another situation is the first digit would be greater than 4 and the second digit would be any even number.
Elements 1:
m - 1 element (4)
n - 4 elements (2,4,6,8)
m x n = 1x4=4
Elements 2:
m - 5 elements (5,6,7,8,9)
n - 5 elements (0,2,4,6,8)
mxn=5x5=25
TOTAL: 25+4=29
SOLVE: Probability
P=29/90
Answer:
goooooooooooooooooood
Step-by-step explanation:and you?
Answer:
Step-by-step explanation:
so we are looking for the x intercept (where the line crosses the x axis).
to do this, we sub in 0 for y and solve for x.
y = 1/2x + 4....sub in 0 for y
0 = 1/2x + 4
-1/2x = 4
x = 4 * -2
x = -8 <=== so it crosses the x axis at (-8,0)
This question is incomplete because it was not written properly
Complete Question
A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)
a) 20%
b) 40%
c) 60%
d) 75%
Answer:
d) 75%
Step-by-step explanation:
We would be solving this question using conditional probability.
Let us represent the percentage of those who passed the first quiz as A = 80%
and
Those who passed the first quiz as B = unknown
Those who passed the first and second quiz as A and B = 60%
The formula for conditional probability is given as
P(B|A) = P(A and B) / P(A)
Where,
P(B|A) = the percent of those who passed the first one passed the second
Hence,
P(B|A) = 60/80
= 0.75
In percent form, 0.75 × 100 = 75%
Therefore, from the calculations above, 75% of those who passed the first quiz to also passed the second quiz.
Implementating the given algorithm in python 3, the greatest common divisors of <em>(</em><em>124</em><em> </em><em>and</em><em> </em><em>244</em><em>)</em><em> </em>and <em>(</em><em>4424</em><em> </em><em>and</em><em> </em><em>2111</em><em>)</em><em> </em>are 4 and 1 respectively.
The program implementation is given below and the output of the sample run is attached.
def gcd(a, b):
<em>#initialize</em><em> </em><em>a</em><em> </em><em>function</em><em> </em><em>named</em><em> </em><em>gcd</em><em> </em><em>which</em><em> </em><em>takes</em><em> </em><em>in</em><em> </em><em>two</em><em> </em><em>parameters</em><em> </em>
if a>b:
<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>b</em>
return gcd (b, a)
<em>#if</em><em> </em><em>true</em><em> </em><em>interchange</em><em> </em><em>the</em><em> </em><em>Parameters</em><em> </em><em>and</em><em> </em><em>Recall</em><em> </em><em>the</em><em> </em><em>function</em><em> </em>
elif a == 0:
return b
elif a == 1:
return 1
elif((a%2 == 0)and(b%2==0)):
<em>#even</em><em> </em><em>numbers</em><em> </em><em>leave</em><em> </em><em>no</em><em> </em><em>remainder</em><em> </em><em>when</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>2</em><em>,</em><em> </em><em>checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>even</em><em> </em>
return 2 * gcd(a/2, b/2)
elif((a%2 !=0) and (b%2==0)):
<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>odd</em><em> </em><em>and</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>even</em><em> </em>
return gcd(a, b/2)
else :
return gcd(a, b-a)
<em>#since</em><em> </em><em>it's</em><em> </em><em>a</em><em> </em><em>recursive</em><em> </em><em>function</em><em>,</em><em> </em><em>it</em><em> </em><em>recalls</em><em> </em><em>the function</em><em> </em><em>with </em><em>new</em><em> </em><em>parameters</em><em> </em><em>until</em><em> </em><em>a</em><em> </em><em>certain</em><em> </em><em>condition</em><em> </em><em>is</em><em> </em><em>satisfied</em><em> </em>
print(gcd(124, 244))
print()
<em>#leaves</em><em> </em><em>a</em><em> </em><em>space</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>first</em><em> </em><em>output</em><em> </em>
print(gcd(4424, 2111))
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