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11 days ago

What key would you use if 10 students chose cartoons

Mathematics

2 answers:

zzz [9K]11 days ago

8 0

Each Box = 2 students. As for the step-by-step explanation, based on the diagram provided, we should identify what key to use for 10 students selecting cartoons. The given key states that Each Box = 3 students. However, using this key makes it impossible to divide 10 students effectively. Hence, it is necessary to adjust the key to a number that is a multiple of 10. Therefore, the key should be Each Box = 2 students. For improved clarity, please refer to figure 1.

lawyer [9.2K]11 days ago

4 0

I'm not entirely certain what you mean, but I'll do my best. Each key corresponds to 3 students, so if 10 opted for cartoons, a more effective key would be 2, 5, and 1. While using 1 could result in clutter, if 10 students chose cartoons and you're applying the 3 key, you might represent 1/3 of a box.

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Last week, the price of apples at a grocery store was $1.60 per pound. This week, apples at the same grocery store are on sale a

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$6.48

Step-by-step explanation:

Initially calculate the 10% discount on $1.60, which results in $1.44. Then multiply $1.44 by 4.5 to find the total, yielding $6.48.

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The profit function P(x), revenue function, R(x), and cost function, C(x), are related by the equation P(x) = R(x) – C(x). Andre

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14 days ago

Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 emp

Zina [9171]

Answer:

a) The likelihood that none of the sampled employees are from the Hawaii plant is 1.74%.

b) The chance that exactly 1 employee from the sample is found working in the Hawaii plant is 8.70%.

c) There is an 89.56% chance that 2 or more employees in the sample are from the Hawaii plant.

d) The probability that 9 employees from the sample are working at the Texas plant is 8.70%.

Step-by-step explanation:

Each employee has two potential employment locations: either Texas or Hawaii. Thus, the binomial probability distribution can be utilized to solve this scenario.

Binomial probability distribution

This distribution defines the probability of achieving exactly x successes in n trials where there are only two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

Here, $C_{n,x}$ denotes the number of ways to choose x objects from a set of n, represented by the subsequent formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of success occurring.

In this context, we know:

The sample comprises 10 employees, therefore $n = 10$ .

a. Calculate the probability that none of the sampled employees are from the Hawaii plant (to 4 decimals)?

Given that 20 out of 60 employees are based in Hawaii:

$p = \frac{20}{60} = 0.333$

We aim to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.333)^{0}.(0.667)^{10} = 0.0174$

Thus, the likelihood that none in the sample are from Hawaii stands at 1.74%.

b. Calculate the probability that 1 employee from the sample is from the Hawaii plant?

This is represented as P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.333)^{1}.(0.667)^{9} = 0.0870$

Therefore, there is an 8.70% possibility that 1 employee in the sample comes from Hawaii.

c. Calculate the probability that 2 or more employees in the sample are from the Hawaii plant?

We can observe two scenarios: either fewer than 2 employees are from Hawaii or 2 and beyond. The combined probabilities equal decimal 1. So:

$P(X < 2) + P(X \geq 2) = 1$

We seek to find $P(X \geq 2)$ .

$P(X \geq 2) = 1 - P(X < 2)$

From problems a and b, we possess values for both probabilities.

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0174 + 0.0870 = 0.1044$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1044 = 0.8956$

Accordingly, the chance that 2 or more employees in this sample operate at the Hawaii plant is 89.56%.

d. Calculate the likelihood that 9 employees in the sample are working at the Texas plant?

This corresponds to the probability found in part b for 1 employee working in Hawaii.

Consequently, there is an 8.70% chance that 9 employees belong to the Texas plant.

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22 days ago

Mr. Deets is making an array to display 9 pictures. For each pair of different factors, there are two arrays he can make. How ma

lawyer [9240]

3 distinct arrangements - 1 x 9 (or 9 x 1) and 3 x 3

Thus, the total number of arrangements = 2 x 2 = 4 = an even count

Initially, we identify pairs of different factors of 9.

The pairs of factors for 9 are:

(1, 9) and (3, 3)

For each of these pairs, Mr. Deets can create 2 arrangements.

This means the total arrangements Mr. Deets can construct = 2 x 2 = 4, confirming that the overall number is even.

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20 days ago

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The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

zzz [9080]

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:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ denotes data points, $\bar{x}$ represents the mean, and n indicates the number of samples.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

The sum of squares of the differences equals 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

The population mean, μ = $30,000

The sample mean, $\bar{x}$ = $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.

$H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}$ A one-tailed t-test will be conducted for this analysis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

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$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

$t_{stat} < t_{critical}$

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

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21 day ago