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2 years ago

When a die is rolled what is the probability of rolling a 2 or a 5?

Mathematics

2 answers:

Travka [436]2 years ago

4 0

$\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{2}{6}=\dfrac{1}{3}$

Elis [28]2 years ago

4 0

1/3 because there is a 1/6 chance of getting any number. So 1/6 + 1/6 = 2/6 which is simplified to 1/3

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Connor have six employees and Bob 41 uniforms for them if you wanted to give them the same number of uniforms so he doesn't have

bagirrra123 [75]

It will be 6 to each person because 6x6 is 36 if u do 6x7 it will be 42 and it said 41 t-shirts not 42

6 0

2 years ago

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For a certain candy, 15% of the pieces are yellow, 1010% are red, 2020% are blue, 55% are green, and the rest are brown

MatroZZZ [7]

A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.

Explanation:
A) The probability that it is brown is the percentage of brown we have. Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%. The probability that one drawn is yellow or blue would be the two percentages added together: 15+20 = 35%. The probability that it is not green would be the percentage of green subtracted from 100: 100-5=95%. Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events. All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
0.5*0.5*0.5 = 0.125 = 12.5%.
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.

The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:

0.85^3 = 0.614 = 61.4%.

The probability that at least one is green is computed by subtracting 1-(probability of no green). We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.

3 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

2 years ago

Stan, a local delivery driver is paid $3.50 per mile driven plus a daily amount of $75. On Monday he is assigned a route that is

Maurinko [17]

3.50m+75 for the 30 miles, he is being paid 105 plus the daily amount of 75 which equals to 180$.

4 0

3 years ago

Two number cubes are rolled.

hichkok12 [17]

Answer:

When two number cubes are rolled,

Sample space= Total Possible outcome {11,22,33,44,55,66,12,21,13,31,14,41,15,51,16,61,23,32,24,42,25,52,26,62,34,43,35,53,36,63,45,54,46,64,56,65}=36

Probability of an event = \frac{\text{Total favorable outcome}}{\text{Total possible outcome}}

Sum of 4 is obtained when ={13,31,22}=3

Probability of getting 4 when 2 number cubes are rolled= \frac{3}{36}=\frac{1}{12}

Sum of the numbers is greater than 9= 10, 11,12={55,64,46,56,65,66}=6

Probability of getting(Sum of the numbers is greater than 9)=\frac{6}{36}=\frac{1}{6}

6 0

3 years ago

Read 2 more answers