A dental laboratory technician earns $14 per hour when she work days and $18 per hour when she works nights. One month she works

Answer:

Step-by-step explanation:

Hello!

The monitoring system alerts the driver when the vehicle's tire pressure drops to 28% below the set target pressure.

Let X be: the target tire pressure for a particular vehicle (measured in pounds per square inch)

a)

X= 28 psi

When the monitoring device alerts at a pressure of 28% below the designated target: X-0.28X

Initially, calculate 28% of 28 psi.

28*0.28= 7.84

Next, subtract this 28% calculation from the target pressure:

28 - 7.84= 20.16

The TPMS will activate its warning at 20.16 psi.

b)

Assuming X~N(μ;σ²)

μ= 28 psi (as the average indicates accurate targeting, this represents the expected tire pressure)

σ= 3 psi

P(X≤20.16)

The standard normal distribution is available in tables. To convert any random variable X with a normal distribution, one subtracts its mean and divides by the standard deviation.

To compute the desired probabilities, the variable value undergoes transformation to fit the standard normal distribution Z, after which standard normal tables are referenced to find probabilities.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now locate the probability corresponding to the Z value using the Z-table. Given the negative result, refer to the left entry; in the first column, find the integer and first decimal for -2.6- and in the first row locate the second decimal for -.-1

The probability link for -2.61 is:

P(Z≤-2.61)= 0.005

c)

The task is to determine the likelihood of encountering a tire at random within the recommended inflation range, expressed as:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Determine both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

<pThe calculated probability of a tire being inflated within the suggested range is 0.498.

I hope this information is useful!

Brent could represent a rectangle with the dimensions of x - 1 and x - 6, then demonstrate that the area equals the combination of x^2, -x, -6x, and 6.

The provided values are

$85 $240 $120 $60 $315 $190 $145 $105 $285 $260 $95 $215 $235 $175 $155

To start, we will arrange these in ascending order

resulting in

$60 $85 $95  $105  $120  $145 $155 $175 $190 $215 $235 $240 $260 $285 $315

The total count is 15

the position of 105 is 4

Now, we can calculate the percentile

Thus, the percentile for the bid of $105 is 26.667%...........Response

Answer:

stratified random sampling method

To calculate this, a specific formula will be necessary. Years = log (total/principal) / [n * log (1 + rate / n)]. Part A) For Calvin: $400 at 5% monthly results in $658.80; Time =? Monthly compounding, n = 12. Thus, Years = log(658.80/400) / [12 * log(1+ (.05/12))]. Subsequently, Years = log( 1.647) / (12 * log(1.0041666667)). Then, Years = 0.21669359917 / 12 * 0.0018058008777. Thus resulting in Years = 0.21669359917 / 0.0216696105. Ultimately, Years ≈ 9.999884362. Part B) For Makayla: $300 at 6% quarterly yields $613.04; Time=? Quarterly compounding, n = 4. Therefore, Years = log(613.04/300) / [4 * log (1 +.06/4)]. This results in Years = log(2.0434666667) / (4 * log(1.015)). Years thus equals 0.31036755784 / (4 * 0.0064660422492), resulting in Years ≈ 11.9999044949. The approximate difference is about 3 years.