At East Middle School, there are 58 left-handed students and 609 right-handed students. The numbers of left- and right-handed st

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!

Let's start by calculating the cost of the first 10 boxes, which totals $75, and the next 10 boxes cost $55.

Together, these 20 boxes amount to $130 spent. With $18 remaining, you can purchase 4 more boxes since 18 divided by 4.5 equals 4.

Therefore, the maximum number of boxes you can buy with $148 is 24.

In the seventh-grade data, the left side appears similar to the right side, unlike in the fifth-grade data. In seventh grade, we can divide the dots into two equal segments, one ranging from 0 to 3 and the other from 4 to 7. The distribution in the first segment is {2, 2, 3, 5}, while the second segment has {5, 3, 3, 1}. These sides mirror each other. When attempting a comparable division in the fifth-grade data, we find one segment from 1 to 4 with a distribution of {2, 3, 1, 4}, and another from 5 to 8 with a distribution of {5, 5, 2, 2}. In this case, the left side does not reflect the right side, indicating a lack of symmetry.

Answer:

Based on the findings, we infer that the average cost of a wedding is lower than the $30,000 that was initially suggested.

Step-by-step explanation:

The following data set is provided: (in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

where denotes data points, represents the mean, and n indicates the number of samples.

The sum of squares of the differences equals 3408750

The population mean, μ = $30,000

The sample mean, = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $697.82

Initially, we formulate the null hypothesis and the alternative hypothesis.

A one-tailed t-test will be conducted for this analysis.

Formula:

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<pnow>

<pgiven the="" situation="">

<pwe do="" not="" accept="" the="" null="" hypothesis="" and="" reject="" it.="">

Thus, we conclude that the average wedding cost is indeed less than the advertised $30,000.

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0.16 + 678% - 4635 + (8886 + 253434 + 77)