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kykrilka
2 months ago
11

What is the greatest common factor of 72 and 112

Mathematics
1 answer:
Zina [12.3K]2 months ago
8 0
112: (2) (2) (2) (2)        (7)
72: (2) (2) (2) 33
__________
GCF: (2) (2) (2)= 8

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Write a difference quotient that best approximates the instantaneous rate of change of g at x=0
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Step-by-step explanation:

The difference quotient represents the slope of the line connecting two points on a curve. To achieve the most accurate estimate, we need to use points that are nearest to x = 0. For this problem, the relevant points are (-0.001, 1.999) and (0.001, 2.001).

m = (2.001 − 1.999) / (0.001 − (-0.001))

m = 1

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2 months ago
The scatterplot below shows the shoe sizes of students and the number of books the students collected for the library book drive
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Respuesta:

B. Generalmente, el tamaño del zapato de un estudiante no influye en la cantidad de libros recolectados.

Explicación paso a paso:

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1 month ago
A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standar
PIT_PIT [12445]

Answer:

The P-value ranges between 2.5% and 5% according to the t-table.

Step-by-step explanation:

A random sample of 16 students from a large university showed an average age of 25 years with a standard deviation of 2 years.

Let \mu = true average age of all students at the university.

So, the Null Hypothesis, H_0 : \mu \leq 24 years {indicating the average age is less than or equal to 24 years}

Alternate Hypothesis, H_A : \mu > 24 years {indicating the average age is significantly greater than 24 years}

Here we employ the One-sample t-test statistics as the population's standard deviation is unknown;

                              T.S. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample average age = 25 years

             s = sample standard deviation = 2 years

             n = sample size = 16

This gives us the test statistics = \frac{25-24}{\frac{2}{\sqrt{16} } } ~ t_1_5

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The value of the t-test statistics is 2.

Moreover, the P-value of the test-statistics can be found as follows;

P-value = P(t_1_5 > 2) = 0.034 {as per the t-table}

Thus, the P-value lies between 2.5% and 5% based on the t-table.

8 0
3 months ago
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