Mathematics, 17.08.2020 01:01 xeno777

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure. Suppose the target tire pressure of a certain car is 29 psi (pounds per square inch.) (a) At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.) (b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi. If the car’s average tire pressure is on target, what is the probability that the TPMS will trigger a warning? (Round your answer to 4 decimal places.) (c) The manufacturer’s recommended correct inflation range is 27 psi to 31 psi. Assume the tires’ average psi is on target. If a tire on the car is inspected at random, what is the probability that the tire’s inflation is within the recommended range? (Round your intermediate calculations and final answer to 4 decimal places.)

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Answer from: cc14ty

At 21.46 psi, the TPMS trigger a warning for this car.

The probability that the TPMS will trigger a warning is 0.0001.

The probability that the tire’s inflation is within the recommended range is 0.6826.

Given that,

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure.

Suppose the target tire pressure of a certain car is 29 psi (pounds per square inch).

We have to determine,

At what psi will the TPMS trigger a warning for this car.

What is the probability that the TPMS will trigger a warning.

What is the probability that the tire’s inflation is within the recommended range.

According to the question,

It is stated that TPMS warns the driver when the tire pressure of the vehicle is 26% below the target pressure.

So, the TPMS trigger a warning for this car when;

Pressure = 29 psi - 26% of 29 psi

$Pressure = 29- (0.26 \times 29) = 21.46psi$

At 21.46 psi, the TPMS trigger a warning for this car.

Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi.

Let X = The pressure at which TPMS will trigger a warning

So, X ~ Normal $\mu = 29, \sigma^2 = 2^2\\$

Now, the probability that the TPMS will trigger a warning is given by = P(X≤ 21.46).

$P(X\leq 21.46) = P (\dfrac{X-\mu}{\sigma}\leq \dfrac{21.46-29}{2}) = P(Z\leq -3.77) = 1-P(Z$

The above probability is calculated by looking at the value of x = 3.77 in the z table which has an area of 0.9999.

The manufacturer’s recommended correct inflation range is 27 psi to 31 psi.

So, the probability that the tire’s inflation is within the recommended range is given by = P(27 psi < X < 31 psi)

P(27psi

Therefore, , P(27 psi < X < 31 psi) = 0.8413 - 0.1587 = 0.6826.

To know more about Probability click the link given below.

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Answer from: abbygreen

(a) At 21.46 psi, the TPMS trigger a warning for this car.

(b) The probability that the TPMS will trigger a warning is 0.0001.

Step-by-step explanation:

We are given that tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure. Suppose the target tire pressure of a certain car is 29 psi (pounds per square inch).

(a) It is stated that TPMS warns the driver when the tire pressure of the vehicle is 26% below the target pressure.

So, the TPMS trigger a warning for this car when;

Pressure = 29 psi - 26% of 29 psi

= $29-(0.26 \times 29)$ = 21.46 psi

At 21.46 psi, the TPMS trigger a warning for this car.

(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi.

Let X = The pressure at which TPMS will trigger a warning

So, X ~ Normal( $\mu=29, \sigma^{2} =2^{2}$ )

Now, the probability that the TPMS will trigger a warning is given by = P(X $\leq$ 21.46)

P(X $\leq$ 21.46) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{21.46-29}{2}$ ) = P(Z $\leq$ -3.77) = 1 - P(Z < 3.77)

= 1 - 0.9999 = 0.0001

The above probability is calculated by looking at the value of x = 3.77 in the z table which has an area of 0.9999.

So, the probability that the tire’s inflation is within the recommended range is given by = P(27 psi < X < 31 psi)

P(27 psi < X < 31 psi) = P(X < 31 psi) - P(X $\leq$ 27 psi)

P(X < 31 psi) = P( $\frac{X-\mu}{\sigma}$ < $\frac{31-29}{2}$ ) = P(Z < 1) = 0.8413

P(X $\leq$ 27 psi) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{27-29}{2}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)

= 1 - 0.8413 = 0.1587

Therefore, P(27 psi < X < 31 psi) = 0.8413 - 0.1587 = 0.6826.

Answer from: Quest

$(4x-101)(2x +3)$

$8x^2 + 12x -202 - 303$

$8x^2+12x-505$

$4x - 101 = 0$

$101/4= 25.25$

$-3/2 = -1.5$

$x = 25.25$ or $x = -1.5$

Answer from: Quest

you need to give us more information since we can't answer

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