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Assoli18 [71]
3 years ago
14

A spinner is divided into 10 equal sections numbered 1 through 10. If the arrow is spun twice, what is the probability the first

number will be even and the second number will be odd?
Mathematics
2 answers:
Sergio039 [100]3 years ago
6 0
Probability of first number being even : 5 even numbers...10 equal sections...probability is 5/10 or 1/2

probability of 2nd number being odd : 5 odd numbers...10 equal sections...so probability of both is 5/10 or 1/2

probability of both occurring : 1/2 * 1/2 = 1/4 <== (or 25% if u need it in percent form or 0.25 if u need it in decimal form)
mylen [45]3 years ago
5 0
\text{p(n%2) = 0 AND p(n%2)} \neq 0
\text{Event 1} = \frac{5}{10} = 0.5
\text{Event 2} = \frac{5}{10} = 0.5
\text{Event 1 * Event 2} = 0.5 * 0.5 = 0.25 \text{
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There are 12 inches in 1 foot. this is equivalent to 60 inches in 5 feet. which proportions can be used to represent this? check
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Answer:12/1=60/5. 1/2=5/60


Step-by-step explanation:

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Two angles are supplements if their sum is 180 degrees. The measurement of
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X represents (smaller) angle 1, y represents (larger) angle 2:

180 = x + y
y = 6x + 5

180 = x + y
180 = x + (6x + 5)
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3 years ago
Mrs. Norton wants to buy health insurance for her family, Plan A costs $620.50 per month. Plan B costs $597.20 per month. How mu
frez [133]

Answer:

Step-by-step explanation:

620.50 * 12 = 7,446

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5 0
3 years ago
Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A1 be the ev
Furkat [3]

Answer:

2.063X10^-4

0.999

0.833

3.26X10^-3

9.85x10^-3

Step-by-step explanation:

B="At least one component needs service during the warranty period"

C="All three components need service during the warranty period"

D="Only the receiver needs service during the warranty period"

E="Exactly one of the three components needs service during the warranty period"

P(A1)=0.0595

P(A2)=0.0598

P(A3)=0.058

a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.0595*0.0598*0.058=2.063X10^-4

b) P(B)=1-P(A1∩A2∩A3)=1-2.063X10^-4=0.999

c) P(C)=P(A1'∩A2'∩A3')=P(A1')P(A2')P(A3')=0.9405*0.9402*0.942=0.833

d) P(D)=P(A1'∩A2∩A3)=P(A1')P(A2)P(A3)=0.9405*0.0598*0.058=3.26X10^-3

e) P(E)=P(A1'∩A2∩A3)+P(A1∩A2'∩A3)+P(A1∩A2∩A3')=3.26X10^-3 + 0.0595*0.9402*0.058+0.0595*0.0598*0.942=9.85x10^-3

6 0
4 years ago
Factorize <br> 4a^2-24a^2b^2+9b^2​
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Hope I can help you

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3 years ago
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