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3 years ago

The object of a popular carnival game is to roll a ball up an incline into regions with different

1 answer:

3 0

The expected value is the mean of the overall observed value or random value. In other words, it is the average of the observed values.

The expected value, E(x) of the given observation is 185

The given parameters can be represented as:

$\begin{array}{cccc}x & {100} & {200} & {300} \ \\ P(x) & {40\%} & {35\%} & {25\%} \ \end{array}$

The following formula calculates the expected value:

$E(x) =\sum x * P(x)$

So, we have:

$E(x) = 100 * 40\% + 200 * 35\% + 300 * 25\%$

$E(x) = 40 + 70 + 75$

$E(x) = 185$

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On Sunday, there were 22 inches of snow. The weather finally began to warm up and by Monday afternoon there were 19 inches. Then

sattari [20]

Answer:

22-3d=x

Step-by-step explanation:

where d is the days that it melts and the amount of snow x is the snow that is left.

3 0

2 years ago

PLEASE HELP!!!! ASAP!!!!!

SCORPION-xisa [38]

-6 5 -3 is the corrdante did those sir

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3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago

Explain the role of the GCF in factoring a polynomial.

Ahat [919]

1. GCF is a factor that is shared by all terms within a polynomial for example, if we have a polynomial of the form A*B + A*C, the GCF is A since both terms have A as a factor 2. after we factor out the GCF, it is placed outside the parentheses, in front of the rest of the polynomial. using our previous example, A*B+A*C = A(B+C) 3. we can verify that our factoring is correct by distributing the GCF to the polynomial within the parentheses

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3 years ago

Find f(x) - g(x) when f(x) = 2x^2 - 4x g(x) = x^2 + 6x

SOVA2 [1]

The last one 3x^ + 2x

7 0

3 years ago