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Sedaia [141]
3 years ago
13

For a t distribution with 16 degrees of freedom, find the area, or probability, in each region. (to 2 decimals)

Mathematics
2 answers:
Fiesta28 [93]3 years ago
6 0

Answer:

The area to the right of 2.120 is 0.0250 rounded to 4 decimal places

Step-by-step explanation:

We are required to find P(t>2.120)

To find this area we have to use the t distribution. Also we can use technology like excel to find the area to the right of 2.120.

The excel function is:

=TDIST(2.120,16,1)=0.0250

Where:

2.120 is the value of the test statistic

16 is the degrees of freedom

1 denotes the one tailed.

Therefore, the area to the right of 2.120 is 0.0250

kirza4 [7]3 years ago
4 0

Answer:

0.03

Step-by-step explanation:

From t-table, at df=16 and t=2.120, probability is 0.025 or 0.03(to 2 decimals).

The probability refers to area to the right of 2.120

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Write an equation for a line on the graph that passes through the points (0.4) and (12,16)
Montano1993 [528]

Answer:

y = x + 4

Step-by-step explanation:

Use the two-point form of the equation of a line.

y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)

y - 4 = \dfrac{16 - 4}{12 - 0}(x - 0)

y - 4 = \dfrac{12}{12}x

y - 4 = x

y = x + 4

3 0
3 years ago
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Marcus reads for 45 minutes per night. He also does chores for 17 minutes per night. In 25 days how many minutes does he spend r
Elis [28]

Answer:

1,125

Step-by-step explanation:

45 x 25 = 1,125

Marcus spends 1,125 minutes more than he does chores.

8 0
3 years ago
We often deal with weighted​ means, in which different data values carry different weights in the calculation of the mean. For​
Veseljchak [2.6K]

Answer:

a) The student's overall average for the class is 87.97.

b) He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.

Step-by-step explanation:

Weighed average:

To solve this question, we find the student's weighed average, multiplying each of his grade by his weights.

Grades and weights:

Scored 82.5 on the midterm, worth 30%.

Scored 88.6 on the final exam, worth 30%.

Scored 91.6 on the homework, worth 40%.

a. On a 100-point scale, what is the student's overall average for the class?

Multiplying each grade by it's weight:

A = 82.5*0.3 + 88.6*0.3 + 91.6*0.4 = 87.97

The student's overall average for the class is 87.97.

b. The student was hoping to get an A in the class, which requires an overall score of 93.5 or higher. Could he have scored high enough on the final exam to get an A in the class?

Score of x on the final class, and verify that the average could be 93.5 or higher. So

A = 82.5*0.3 + 88.6*0.3 + 0.4x

A \geq 93.5

82.5*0.3 + 88.6*0.3 + 0.4x \geq 93.5

0.4x \geq 93.5 - (82.5*0.3 + 88.6*0.3)

x \geq \frac{93.5 - (82.5*0.3 + 88.6*0.3)}{0.4}

x \geq 105.425

He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.

3 0
3 years ago
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Liono4ka [1.6K]
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Refer to the qestion below EXPLAIN (f * g * h) (x)
erastovalidia [21]
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Should get C for an answer.
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