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salantis [7]
2 years ago
8

Find the measure of each numbered angle !! (You don’t have to answer for all of them, I just need at least one )

Mathematics
2 answers:
olga2289 [7]2 years ago
7 0

Answer:

7) 123°

8) 38°

9) 68°

10) 65°

11) 113°

12) 52°

Natalka [10]2 years ago
5 0

Answer:

7. 123

8. 38

9. 68

10. 65

11. 113

12. 52

Step-by-step explanation:

You might be interested in
The total weight of 30 bags of flour and 4 bags of sugar is 42.6 kg. If each bag of sugar weighs 0.75 kg, what is the weight of
alina1380 [7]
The answer is: Each bag of flour weighs 1,32 kg. 

If we the total weight of the bags is given, and we know both the number of bags of flour and sugar, and we also know the weight of each bag of sugar, then we have to find the unknown, which is X. 30 times X plus 4 times the weight of a bag of sugar would equal 42.6kg. Next step is to put the unknown on one side and the known values on the other side. We have 30 times X equals 42.6 minus 4 times 0.75. 
To find X we need to divide the value with 30, or to sum up 

30X + 4*0.75 = 42.6
30X + 3 = 42.6
X = (42.6 - 3) / 30
X = 1.32 kg
7 0
3 years ago
Read 2 more answers
Mr.Johnson weighs 135 pounds. Her husband weighs 80 kilogram. How many fewer pounds does Ms.Johnson weigh than her husband?
sweet-ann [11.9K]
First you convert kilograms to pounds,
1 kilogram = 2.205 pounds
next you times the number of pounds per kilogram by the number of kilograms given that need to be turned into pounds.
80 times 2.205 = 176.4
next you subtract to get the answer
176.4 - 135= 41.4
so Ms. Johnson weighs 41.4 pounds less than herr husband


3 0
3 years ago
Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
3 years ago
107=5x+17 simplify your answer as much as possible
olga nikolaevna [1]
107-7=90
90divided 5=18
18*5=90+17=107
I did the work ♥♥♥♥
6 0
3 years ago
Read 2 more answers
Craig has of collection of 220 marbles. He lets his friend have 14 of his collections and sells 15 of his collection to the nove
Illusion [34]
220-14= 206
206-15=191
191-15=176
Answer: 176
3 0
3 years ago
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