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Verdich [7]
3 years ago
9

PLS HELP I WILL MARK YOU BRAINLIEST!!!! The table represents some points on the graph of an exponential function.

Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0
<h3>Answer:  Choice B</h3><h3>f(x) = 18(6/5)^x</h3>

====================================================

Explanation:

When x = 0, we have y = 18. The y intercept value is 18. This is the initial value, and it's the value of 'a' in the equation below

y = a*b^x

The value of b is tied to the growth or decay rate, depending on if b > 1 or if 0 < b < 1 respectively.

In this case, we have growth due to y increasing as x increases. They increase together.

When x increases by 1 like this, the value of b can be determined by dividing any y term over its previous value. So for instance, we can say b = 21.6/18 = 1.2 or we could say b = 25.92/21.6 = 1.2

Then note how 1.2 = 6/5, which means b = 6/5

Therefore, y = a*b^x becomes y = 18(6/5)^x

--------------

We can confirm this equation by plugging in x values from the table and we should get the y values, or f(x) values, listed

Example: plug in x = 0 to find that

y = 18(6/5)^x

y = 18(6/5)^0

y = 18(1)

y = 18 which matches

Then try x = 1

y = 18(6/5)^x

y = 18(6/5)^1

y = 18(1.2)

y = 21.6 that also matches

I'll let you confirm the other x values.

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Answer:

AC = 8x - 4

Step-by-step explanation:

AC = AB + BC

AC = 6x - 7 + 2x + 3

AC = 6x + 2x - 7 + 3

AC = 8x - 4

3 0
3 years ago
In a certain manufacturing process, it is known that, on average, 1 in every 100 items is defective.
Rashid [163]

Answer:

A. 0.009899

B. 0.005624

Step-by-step explanation:

Data:

Let the probability that an item is defective = \frac{1}{100}

The probability that the item is not defective = \frac{99}{100}

The probability that the fifth item is defective = \frac{1}{100}* \frac{98}{99}

                                                                            = 0.009899

Probability that one in 5 items is defective  = 0.005624

5 0
4 years ago
Which expression forms an equation with the given expression?
harina [27]

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6 0
4 years ago
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Alexxx [7]
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8 0
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.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(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 <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(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 <em>X</em> = all 3 vehicle passes and <em>Y</em> = 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
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