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alina1380 [7]
3 years ago
13

Kamila has two number cubes each labeled 1 to 6. She is going to conduct an experiment by tossing both cubes a total of 150 time

s. She will find the sum of the two numbers in each role
a. Possible outcomes? __________
b. Probability of tossing a sum of six? _________
c. How many times should Kamila toss a sum of 7? _________
d. How many times should Kamila toss a sum of 10 or greater? _______
Mathematics
2 answers:
jasenka [17]3 years ago
7 0
A) There are 36 possible outcomes.
b) The probability of a sum of 6 is 5/36.
c) She should roll a sum of 7 45 times.
d) She should roll a sum of 10 45 times.

Explanation
a) There are 6 outcomes for the first die and 6 outcomes for the second one.  By the fundamental counting principle, there are 6*6 = 36 outcomes for both dice together.

b) The ways to get a sum of 6 are:
1&5; 2&4; 3&3; 4&2; 5&1.  There are 5 possibilities out of a total of 36, or 5/36.

c) The ways to get a sum of 7 are:
1&6; 2&5; 3&4; 4&3; 5&2; 6&1.  There are 6 out of 36, or 6/36=1/6.  Since she is rolling the dice 150 times, she should get a sum of 6
1/6(150) = 150/6 = 45 times.

d) The ways to get a sum of 10 or more are:
4&6; 5&5; 6&4; 5&6; 6&5; 6&6

There are 6 ways out of 36, or 6/36 = 1/6.  Since she is rolling the dice 150 times, she should get a sum of 10 or more
1/6(150) = 150/6 = 45 times.
bonufazy [111]3 years ago
4 0

Answer:

Step-by-step explanation:

a) When two number cubes are thrown possible outcomes are

6x6  =36

b) Probability of tossing a sum of 6

Favourable outcomes = (1,5)(5,1)(2,4) (4,2) (3,3)  

Hence probability =\frac{5}{36}

c) For getting 7 favourable outcomes are

(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)

i.e one out of 6 chances.

Hence in 150 times Kamala should toss a sum of 7 ---  \frac{1}{6} *150 =25

d) Favourable outcomes for sum of 10 or greater

= (4,6) (6,4) (5,5) (5,6)(6,5)(6,6)

Probability for getting sum of 10 or greater = \frac{6}{36} =\frac{1}{6}

No of times Kamila should toss 10 or greater

= \frac{150}{6} =25

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\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}

and the y-coordinate (S-N axis) is:

\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}

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Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}

The angle, from E to N, can be calculated as:

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If we want to express it from N to E, we substract the angle from 90°:

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Answer: the final location can be represented with the vector (282.53, 331.59).

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2) the direction is N-40°-E.

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