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Nutka1998 [239]
3 years ago
14

Saby used a probability simulator to roll a 6-sided number cube and flip a coin 100 times. The results are shown in the tables b

elow: Number on the Cube Number of Times Rolled 1 18 2 25 3 12 4 32 5 3 6 10 Heads Tails 56 44 Using Saby's simulation, what is the probability of rolling a 4 on the number cube and the coin landing heads up?
Mathematics
1 answer:
Degger [83]3 years ago
8 0
Short Answers
P(Both) = 0.1792
Givens
The total number of roles = 100
Total number of heads = 56
Number of times rolled a four =  32 [check this for me].

Set up the probabilities.
P(H) = 56/100
P(4) = 32/100
P(H)*P(4) = P(Both)

Substitute and solve
P(Both) = 56/100 * 32/100 = 1792/10000
P(Both) = 112/625 because there are many cancellations. Or you could just express it as 0.1972

You might be interested in
12x + 4 divided by 4 plus 20x + 5 divided by 5
Rasek [7]
12x + 4 ➗ 4 + 20x + 5 ➗ 5

1. PEMDAS (order of operations) solve the division problems

(4 divided by 4 & 5 divided by 5)
12x + 1 + 20x + 1

2. Swap around the order of operations as commutative property states that you can swap the order of an addition problem and get the same answer.

12x + 20x + 1 + 1

3. Add like terms

32x + 2

This is the most simplified answer you could get as you can not add 32x and 2 as they are not like terms.
6 0
2 years ago
How many zeroes do we write when we write all the integers 1 to 243 in base 3?
Monica [59]

Answer:

289 numbers

Step-by-step explanation:

Above you will find the list of integers from 1 to 243 in base 3:

(1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1022, 1100, 1101, 1102, 1110, 1111, 1112, 1120, 1121, 1122, 1200, 1201, 1202, 1210, 1211, 1212, 1220, 1221, 1222, 2000, 2001, 2002, 2010, 2011, 2012, 2020, 2021, 2022, 2100, 2101, 2102, 2110, 2111, 2112, 2120, 2121, 2122, 2200, 2201, 2202, 2210, 2211, 2212, 2220, 2221, 2222, 10000, 10001, 10002, 10010, 10011, 10012, 10020, 10021, 10022, 10100, 10101, 10102, 10110, 10111, 10112, 10120, 10121, 10122, 10200, 10201, 10202, 10210, 10211, 10212, 10220, 10221, 10222, 11000, 11001, 11002, 11010, 11011, 11012, 11020, 11021, 11022, 11100, 11101, 11102, 11110, 11111, 11112, 11120, 11121, 11122, 11200, 11201, 11202, 11210, 11211, 11212, 11220, 11221, 11222, 12000, 12001, 12002, 12010, 12011, 12012, 12020, 12021, 12022, 12100, 12101, 12102, 12110, 12111, 12112, 12120, 12121, 12122, 12200, 12201, 12202, 12210, 12211, 12212, 12220, 12221, 12222, 20000, 20001, 20002, 20010, 20011, 20012, 20020, 20021, 20022, 20100, 20101, 20102, 20110, 20111, 20112, 20120, 20121, 20122, 20200, 20201, 20202, 20210, 20211, 20212, 20220, 20221, 20222, 21000, 21001, 21002, 21010, 21011, 21012, 21020, 21021, 21022, 21100, 21101, 21102, 21110, 21111, 21112, 21120, 21121, 21122, 21200, 21201, 21202, 21210, 21211, 21212, 21220, 21221, 21222, 22000, 22001, 22002, 22010, 22011, 22012, 22020, 22021, 22022, 22100, 22101, 22102, 22110, 22111, 22112, 22120, 22121, 22122, 22200, 22201, 22202, 22210, 22211, 22212, 22220, 22221, 22222, 100000)

If you count them, you will find that there are 289 numbers in total!

8 0
3 years ago
A coin is tossed 5 times and a tail is the outcome each time. what is the probability of getting a tail on the sixth throw
Norma-Jean [14]
I believe it would be
1/5
5 0
3 years ago
What is the average of the four smallest of integers ?
Elanso [62]
This question is easier than it seems. If there are 9 consecutive integers with the mean being 16, then 16 must be the median. So just add four on top and four below 16 to give you your range of 9 values, 12-20. 12+13+14+15+16+17+18+19+20=144, 144/9=16. So the mean of the first four numbers is (12+13+14+15)/4= 13.5. If you check, 13.5 is also the median of these four numbers.

 
6 0
3 years ago
All we asked for the question not all that
maria [59]

Answer:

...what?

Step-by-step explanation:

≥≧≦≤

.^◡^.

<em>aM gObLiN gImMiE yE pOiNtS!    </em>            ( thanks )

8 0
3 years ago
Read 2 more answers
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