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allochka39001 [22]
2 years ago
13

E saca una bola de billar al azar de una caja que contiene 15 bolas de billar numeradas del 1 al 15 y se registra el número obte

nido con la letra "n". Encuentra la probabilidad de que el número obtenido "n" de la bola de billar sacada de la caja sea mayor a 10 y "n" sea par.
Mathematics
1 answer:
Julli [10]2 years ago
5 0

Answer:

P = 2/7 = 0.2857 or 28.57%

Step-by-step explanation:

First, we know that we have 15 balls, and we need to know which one of them is pair, and higher than 10.

So, we first need to calculate the probability that the obtained number is pair.

The pair numbers are 2,4,6,8,10,12,14

We have 7 numbers out of 15 ---> P(B) = 7/15

From those numbers, only two of them are higher than 10, which are 12 and 14, so: P(A|B) = 2/15

To get the probability to get a number higher than 10 and pair:

P(A/B) = P(A|B) / P(B)

P(A/B) = 2/15 / 7/15 = 2/7 = 0.2857

To get the percentage: 0.2857 * 100 = 28.57%.

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a(n) = a + b(n - 1)
a(n) = 9 + 5(n - 1)

The answer is B
8 0
2 years ago
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A new car is purchased for 15300 dollars. The value of the car depreciates at 14.25% per year. What will the value of the car be
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Answer:

\$6,082.70  

Step-by-step explanation:

we know that

The  formula to calculate the depreciated value  is equal to  

V=P(1-r)^{x}  

where  

V is the depreciated value  

P is the original value  

r is the rate of depreciation  in decimal  

x  is Number of Time Periods  

in this problem we have  

P=\$15,300\\r=14.25\%=0.1425\\x=6\ years

substitute in the formula

V=15,300(1-0.1425)^{6}  

V=15,300(0.8575)^{6}  

V=\$6,082.70  

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2 years ago
Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla
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Answer:

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Step-by-step explanation:

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Noah wants to put $1,000 in a savings account with a 1.5% annual interest rate. How much more money will he have after one year
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Assuming that the 1.5% annual interest is split into monthly basis with the same amount, then the monthly interest should be: 1.5%/12= 0.125%.
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If you put $1000 for monthly interest, the saving account would become: ($1000*(100%+0.125%)= $1000*1.0151035559= $1015.10

Then, the money difference should be:  $1015.10-$1015= $0.10
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2 years ago
Without properly working spark plugs, a vehicle will not run. For a specific vehicle, the spark plugs are supposed to have a gap
Salsk061 [2.6K]

Answer:

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3)  = 0.9922</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the mean of the Normal distribution = 4.1mm</em>

<em>Given that the standard deviation of the Normal distribution = 0.0075mm</em>

<em>Let 'X' be the random variable in a normal distribution</em>

Given that X₁ = 3.9mm

Z_{1} = \frac{X_{1} -mean}{S.D}

Z_{1} = \frac{3.9 -4.1}{0.075} =  -2.666

Given that X₂ = 4.3mm

Z_{2} = \frac{X_{2} -mean}{S.D}

Z_{1} = \frac{4.3 -4.1}{0.075} =  2.666

<u><em>Step(ii):-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3) = P(-2.666≤Z≤2.666)</em>

<em>                      = P(Z≤2.666)-P(Z≤-2.666)</em>

<em>                     = 0.5 +A(2.666) - (0.5-A(2.666)</em>

<em>                    = 2 × A(2.666)</em>

<em>                  = 2×0.4961</em>

<em>                  = 0.9922</em>

<u><em>Final answer:-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3)  = 0.9922</em>

<em></em>

3 0
1 year ago
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