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3 years ago

The probability of getting a 6 after rolling a fair dice 16.

Mathematics

1 answer:

Butoxors [25]3 years ago

4 0

Answer:

P(6) = 5

Step-by-step explanation:

The dice is cast 30 times (k = 30)

So if the big numbers apply, then you would expect the outcome to be evenly distributed, meaning

5 x 1

5 x 2

5 x 3

5 x 4

5 x 5

5 x 6

____ +

30 casts in total (evenly distributed over all numbers of the dice)

P(6) = 1/6 * 30

P(6) = 5

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Meg saved coins she found for a year. She found a total of 95 pennies, 13 nickels, 41 dimes, and 11 quarters. She would like to

Katen [24]

Answer:

40 coins in each piggy bank

Step-by-step explanation:

Add all the coins together.

95 pennies+ 13 nickels+ 41 dimes+ 11 quarters=160 coins

Divide the number of coins by 4.

160 coins divided by 4 piggy banks= 40 coins in each piggy bank

I hope this helps ;)

5 0

3 years ago

Solve for x round to the nearest tenth

lidiya [134]

Answer:

47.0

Step-by-step explanation:

In this right angle triangle, we are faced with a challenge of two sides. The opposite side and the adjacent side, hence the tangent is used.

Where it is the opposite side and hypothenus side, the sine is used and when it is the hypothenus side and adjacent side, the cosine is used.

Hence, we have tan62°=x/25

We cross multiply, to have

25(tan 62°)= x

x = 47.01816

In rounding up numbers, number 1 to 4 will be rounded up to zero, while numbers 5 to 9 will be rounded up to 1.

Rounding up 47.01816 to the nearest tenth. The tenth value is the figure is 0, before it we have 1, which is to hundredth. 1 will be rounded up to zero.

So we have 47.0

3 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

John purchased a carr in 1961 for $12,000. Experts estimate that its value will increase by 5% per year. How much will be the co

dexar [7]

Answer:

Step-by-step explanation:

I'm super lazy rn and it doesn't need explanation

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3 years ago

Kaitlyn is walking on a treadmill at a constant pace for 28 minutes. She has programmed the treadmill for a 2-mile walk. The dis

maw [93]

Is this multiplying or what??? Anyways try calculating those numbers so basically you need to calculate 28x2 or if you have to divide them or multiply them you would get the answer is very simple

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2 years ago