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kicyunya [14]
3 years ago
12

Three dice are going to be rolled. What is the probability that the first die rolled is a 1, the second die rolled is a 2, and t

he third die rolled is a 3? (If necessary, round to the nearest tenth)
Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
Answer: The probability is 0.46%.

The chance of each given event happening is 1/6 because there are 6 different number on the dice and only 1 number is chose.

Therefore to find the combined probability, we have to multiply all the individual probabilities.

(1/6) x (1/6) x (1/6)

Or

(1/6)^3

The answer is about 0.46%,
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MissTica

Answer:

-14

Step-by-step explanation:

-7*2= -14

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The bike store marks up the wholesale cost of all of the bikes they sell by 30%.
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Answer:

Suppose that the wholesale of a bike is A (A is the 100% in this case). If we have an increase of 30% for the sale price, we have the new price of:

Price = A + (30%/100%)A = A + 0.3*A = (1.3)*A

Then if we know that the price tag of the bike is $125, then we have:

$125 = (1.3)*A

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The wholesale cost of the bike is $96.15

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50 POINTS<br><br> What is the volume of this?
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The volume is 30.

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 If the trapezoid below is reflected across the x-axis, what are the coordinates of B'?
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The x-coordinate remains the same as the x-coordinate of point B.
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A jar of peanut butter contains 454 g with a standard deviation of 10.2 g. Find the probability that a jar contains more than 46
Fofino [41]

Answer:

The probability that a jar contains more than 466 g is 0.119.

Step-by-step explanation:

We are given that a jar of peanut butter contains a mean of 454 g with a standard deviation of 10.2 g.

Let X = <u><em>Amount of peanut butter in a jar</em></u>

The z-score probability distribution for the normal distribution is given by;

                                Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 454 g

           \sigma = standard deviation = 10.2 g

So, X ~ Normal(\mu=454 , \sigma^{2} = 10.2^{2})

Now, the probability that a jar contains more than 466 g is given by = P(X > 466 g)

            P(X > 466 g) = P( \frac{X-\mu}{\sigma} > \frac{466-454}{10.2} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                                  = 1 - 0.881 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

4 0
3 years ago
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