How dot figure out the perimeter of Question 2, a

51.96 mm

Step-by-step explanation:

Perimeter of circle = 2 × pi × r

Perimeter of small circle

= 2 × 3.14 × 10/2

= 31.4 mm

half of perimeter of small circle = 31.4/2 = 15.7 mm

Perimeter of large circle

= 2 × 3.14 × (5+4)

= 2 × 3.14 × 9

= 56.52

half of perimeter of large circle = 56.52/2 = 28.26 mm

All line = half line small circle + half line large circle + (4 × 2)

All line = 15.7 + 28.26 + 8

= 51.96 mm

It does not appear that, as a group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

where are data points, is the mean and n is the number of observations.  

Sum of squares of differences = 4739.733

Population mean, μ = 60 minutes

Sample mean, = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

We use Two-tailed t test to perform this hypothesis.

Formula:

Putting all the values, we have

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear that, as a group, the students are reasonably good at estimating one minute.