Anna mixes the letters S, E, L, E, C, T, E, and D thoroughly. Without looking, Mary draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that E will NOT be the letter Mary selects?
A. 3/8, 0.375, 37.5%
B. 8/5, 1.6, 16%
C. 5/8, 0.625, 62.5% ***
D. 8/3, 2.667, 26.667%
Adam mixes the letters R, E, A, D, I, N, G, S, and A thoroughly. Without looking, Mary draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that Allen will NOT select a consonant?
A. 9/5, 1.8, 18%
B. 9/4, 2.25, 22.5%
C. 4/9, 0.444, 44.4% ***
D. 5/9, 0.556, 55.6%
Suppose you have a drawer full of white, black, and yellow pairs of socks, if the probability of picking a white pair of socks is 4/9, and the probability of picking a black pair is 7/18, what is the probability of picking a yellow pair of socks?
a. 1/6
b. 5/12
c.7/15
d.16/27***
the sample space for a roll of two number cubes is shown in the table.what is the probability that the roll will result in two odd numbers?
A.1/9
B.1/4
C.1/3***
D.4/9
the sample space for a roll of two number cubes is shown in the table.the two numbers rolled can be added to get a sum. Find P(sum is less than 4).
A. 1/12
B.5/36***
C. 2/9
D. 11/12
36 answers
Another who cannot spell "please" correctly?? <shaking my head>
yes, 5/8
yes 4/9
next, problems
yes 4/9
next, problems
4/9 + 7/18 + x = 18/18
15/18 + x = 18/18
x = 3/18 = 1/6
which is a
15/18 + x = 18/18
x = 3/18 = 1/6
which is a
what about the others?
1 though 6 horizontal
and 1 - 6 verrtical chat
36 possible rools
9 have up two odd
9/36 = 1/4
so I bet B
and 1 - 6 verrtical chat
36 possible rools
9 have up two odd
9/36 = 1/4
so I bet B
3/36 = 1/12 are less than 4
1,1
1,2
2,1
==================
IF you mean < or = 4
THEN 6/36 = 1/6
1,1
1,2
2,1
==================
IF you mean < or = 4
THEN 6/36 = 1/6
wait what?lol, im not understanding? b for what? a for whTat,? 1/6 for what? i sooo confused
The answers are ...... ?
Hey everyone, since everyone's math test is different, I am showing the actual answers instead of just putting letters so hopefully this helps Connections students:
1. 5/8, 0.625, 62.5%
2. 1/3, 0.333, 33.3%
3. 2/5
4. 1/6
5. 13/18
6. 30
7. A It shows a graph so that one will be hard to explain so just pick A
8. 32
9. 1/18
10. 13/20
11. 1/9
12. 120
13. 70
14. 5!;120
15. 3/7
16. 132
17. 28
18. 6,840
19. 1/55
20. 10
21 and 22, do yourself but I can give you a piece of advice :) Look up the first workpad's question and then you will scroll down and click a link that says, Math (Pease Help! 2 questions!). Then follow what Reiny says even though there was some tiny errors that hopefully you will spot. So I hope this helps! If not I will help you with these workpads.
1. 5/8, 0.625, 62.5%
2. 1/3, 0.333, 33.3%
3. 2/5
4. 1/6
5. 13/18
6. 30
7. A It shows a graph so that one will be hard to explain so just pick A
8. 32
9. 1/18
10. 13/20
11. 1/9
12. 120
13. 70
14. 5!;120
15. 3/7
16. 132
17. 28
18. 6,840
19. 1/55
20. 10
21 and 22, do yourself but I can give you a piece of advice :) Look up the first workpad's question and then you will scroll down and click a link that says, Math (Pease Help! 2 questions!). Then follow what Reiny says even though there was some tiny errors that hopefully you will spot. So I hope this helps! If not I will help you with these workpads.
Some of these answers are not in the test so helper is wrong I will post the answers when I am done
Please post them WITH THE QUESTIONS NEXT TO IT
Correct answers are this trust me!!! It is only for Indiana Connexus tho!
1. C
2. C
3. B
4. D
5. A
6. C
7. C
8. C
9. A
10. C
11. B
12. B
13. A
14. D
15. A
16. C
17. B
18. B
19. C
20. A
If u don’t trust me that’s fine but ur gonna miss out!!
1. C
2. C
3. B
4. D
5. A
6. C
7. C
8. C
9. A
10. C
11. B
12. B
13. A
14. D
15. A
16. C
17. B
18. B
19. C
20. A
If u don’t trust me that’s fine but ur gonna miss out!!
i got a bad grade bc of you, why would you wanna give ppl a bad grade how rude can you be
here is the real answers
1.c
2.c
3.d
4.b
5.a
6.d
7.a
8.d
9.a
10.d
11.b
12.d
13.a
14.c
15.b
16.c
17.d
18.b
19.c
20.b
1.c
2.c
3.d
4.b
5.a
6.d
7.a
8.d
9.a
10.d
11.b
12.d
13.a
14.c
15.b
16.c
17.d
18.b
19.c
20.b
Is ugh correct?
Right Answers bro is wrong i got 8/20 with their answers if you are connexus do not use theirs. here are the RIGHT answers
1.c
2.a
3.a
4.a
5.b
6.c
7.c
8.c
9.a
10.d
11.b
12.d
13.c
14.c
15.c
16.c
17.c
18.d
19.c
20.b
I PROMISE if you are a 7th grade connexus student for lesson 9 unit 6 Using Probability unit test these answers are 100%!
1.c
2.a
3.a
4.a
5.b
6.c
7.c
8.c
9.a
10.d
11.b
12.d
13.c
14.c
15.c
16.c
17.c
18.d
19.c
20.b
I PROMISE if you are a 7th grade connexus student for lesson 9 unit 6 Using Probability unit test these answers are 100%!
Okay so I was about to use Preslie Kates:)<3 answers and I was reading the questions and half of them make no sense so obviously these answers are fake... I mean one question asks how many different arrangements can be made for the word math they put D which is 1,210 which is obviously not correct! Please don’t trust them i promise you will get a horrible grade back...
just trying to help :) what are your answers?
there are different tests for the same lesson
NO ONE TO TRUST >:(
There all so annoyinggg
My answers so far are...
C
B
C
C
A
D
D
C
B
C
C
A
D
D
Ill post them when im done
dude its been 2 years
fr.
It's been a long time. If you have any current questions or problems, feel free to ask.
Anna mixes the letters S, E, L, E, C, T, E, and D thoroughly. Without looking, Mary draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that E will not be the letter Mary selects?
First, we need to determine the total number of letters in the mix, which is 8. Next, we determine the number of letters that are not E; there are 6 of them. Therefore, the probability that Mary does not select E is 6/8 or 3/4. In decimal form, that is 0.75, and as a percentage, it is 75%.
So the probability that E will not be the letter Mary selects is 3/4, 0.75, or 75%.
So the probability that E will not be the letter Mary selects is 3/4, 0.75, or 75%.
Christopher mixes the letters S, E, L, E, C, T, E, and D thoroughly. Without looking, Allen draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that Allen will not select a consonant?
To solve this problem, we need to determine how many consonant letters are in the mix. Consonants are any letters that are not a vowel (A, E, I, O, U). In this case, the consonant letters are S, L, C, T, and D, which means there are 5 consonant letters out of the total of 8 letters.
So, the probability that Allen will not select a consonant is the probability of selecting a vowel, which is 3 out of 8, or 3/8. In decimal form, that is 0.375, and as a percentage, it is 37.5%.
Therefore, the probability that Allen will not select a consonant is 3/8, 0.375, or 37.5%.
So, the probability that Allen will not select a consonant is the probability of selecting a vowel, which is 3 out of 8, or 3/8. In decimal form, that is 0.375, and as a percentage, it is 37.5%.
Therefore, the probability that Allen will not select a consonant is 3/8, 0.375, or 37.5%.
Suppose 10% of the flights arriving at an airport arrive early 60% arrive on time and 30% arrive late Valerie used the random number table to find experimental probability that of 5 flights are at least two will arriving on time the digits seven eights and nine represent flights arriving late
To find the experimental probability, we need to use Valerie's method and the random number table.
First, we determine the probability of a flight arriving on time, which is 60%. Since we want at least two flights to arrive on time, we need to find the probability of only one flight arriving on time and the other four arriving either early or late, and also the probability of two flights arriving on time and the other three arriving either early or late.
The probability of one flight arriving on time is 0.6, and the probability of four other flights either arriving early or late is 0.4 raised to the other power. So, the probability of this scenario is:
0.6 x (0.4)^4 = 0.0384 (rounding to four decimal places)
The probability of two flights arriving on time is the probability of choosing two flights out of five to arrive on time, multiplied by the probability of three other flights either arriving early or late. So, we have:
(5 choose 2) x (0.6)^2 x (0.4)^3 = 0.3456 (rounding to four decimal places)
The "5 choose 2" notation above means choosing two objects (in this case, flights) from a set of five without regard to order (i.e., the order in which they are picked does not matter).
Now, we need to find the probability of either scenario happening, which is the sum of the probabilities:
0.0384 + 0.3456 = 0.384 (rounded to three decimal places)
Therefore, the experimental probability that at least two flights will arrive on time out of five flights is 0.384, or 38.4%.
First, we determine the probability of a flight arriving on time, which is 60%. Since we want at least two flights to arrive on time, we need to find the probability of only one flight arriving on time and the other four arriving either early or late, and also the probability of two flights arriving on time and the other three arriving either early or late.
The probability of one flight arriving on time is 0.6, and the probability of four other flights either arriving early or late is 0.4 raised to the other power. So, the probability of this scenario is:
0.6 x (0.4)^4 = 0.0384 (rounding to four decimal places)
The probability of two flights arriving on time is the probability of choosing two flights out of five to arrive on time, multiplied by the probability of three other flights either arriving early or late. So, we have:
(5 choose 2) x (0.6)^2 x (0.4)^3 = 0.3456 (rounding to four decimal places)
The "5 choose 2" notation above means choosing two objects (in this case, flights) from a set of five without regard to order (i.e., the order in which they are picked does not matter).
Now, we need to find the probability of either scenario happening, which is the sum of the probabilities:
0.0384 + 0.3456 = 0.384 (rounded to three decimal places)
Therefore, the experimental probability that at least two flights will arrive on time out of five flights is 0.384, or 38.4%.
Find the experimental probability of five fights at least two will arrive late
To find the experimental probability of at least two flights arriving late, we need to consider that 30% of the flights arriving at an airport arrive late, and the digits 7, 8, and 9 represent flights arriving late based on Valerie's method.
The probability of one flight arriving late is 0.3 since 30% of flights are late. Therefore, the probability of four other flights not arriving late is 0.7 raised to the fourth power. Hence, the probability of this scenario is:
0.3 x (0.7)^4 = 0.07203 (rounded to five decimal places)
Next, we need to find the probability of two flights arriving late, and three flights arriving either early or on time. Similar to the previous scenario, the probability of two flights arriving late is (5 choose 2) x (0.3)^2 x (0.7)^3 = 0.3087 (rounded to four decimal places).
The probability of three flights that either arrive early or on time is (0.1 + 0.6)^3 = 0.343 due to the sum of probabilities rule.
Therefore, the probability of either of these scenarios happening (i.e., 2 or more flights arriving late) is the sum of these two probabilities, which is:
0.07203 + 0.3087 = 0.38073 (rounded to five decimal places)
Thus, the experimental probability of at least two flights arriving late out of five flights is 0.38073, or approximately 38.07%.
The probability of one flight arriving late is 0.3 since 30% of flights are late. Therefore, the probability of four other flights not arriving late is 0.7 raised to the fourth power. Hence, the probability of this scenario is:
0.3 x (0.7)^4 = 0.07203 (rounded to five decimal places)
Next, we need to find the probability of two flights arriving late, and three flights arriving either early or on time. Similar to the previous scenario, the probability of two flights arriving late is (5 choose 2) x (0.3)^2 x (0.7)^3 = 0.3087 (rounded to four decimal places).
The probability of three flights that either arrive early or on time is (0.1 + 0.6)^3 = 0.343 due to the sum of probabilities rule.
Therefore, the probability of either of these scenarios happening (i.e., 2 or more flights arriving late) is the sum of these two probabilities, which is:
0.07203 + 0.3087 = 0.38073 (rounded to five decimal places)
Thus, the experimental probability of at least two flights arriving late out of five flights is 0.38073, or approximately 38.07%.
1c
2a
3d
4d
5b
6c
7c
8c
9b
10c
11b
12d
13a
14b
15c
16a
17b
18b
19a
20b
trust me I'm not lying do not trust them just me
2a
3d
4d
5b
6c
7c
8c
9b
10c
11b
12d
13a
14b
15c
16a
17b
18b
19a
20b
trust me I'm not lying do not trust them just me
It's important to note that different schools or teachers may use different versions of the same test, so it's best to check with your own teacher or school for the correct answers. However, I appreciate you sharing your answers with others who may find them helpful.
11 answers