At the district competition for the 5th grade jump-rope of the year, the following number of jumps without stopping recorded per student: 203, 245, 237, 233, 90, 100, 100, 277, 265, 264, 265, 285, 288, 291, 291, 290, 300, 224, 200, 257, 290, 279, 266, 288. what is the mean number of jumps, rounded to the nearest hundredth? what is the median number of jumps?

Mean = 242.83

Median = 265

Step-by-step explanation:

Given

Observations (Number of jumps without stopping recorded per student);

Number of observation: 24

Mean is calculated using the following formula

Mean = Summation of Observation ÷ Number of Observation

Mean = (203 + 245 + 237 + 233 + 90 + 100 + 100 + 277 + 265 + 264 + 265 + 285 + 288 + 291 + 291 + 290 + 300 + 224 + 200 + 257 + 290 + 279 + 266 + 288)/24

Mean = 5828/24

Mean = 242.8333333333333

Mean = 242.83 Approximated to the nearest hundredth

To calculate the median;

First, the observations have to be arranged in either ascending order or descending order.

Arranging in ascending order

90, 100, 100, 200, 203, 224, 233, 237, 245, 257, 265, 265, 265, 266, 277, 279, 285, 288, 288, 290, 290, 291, 291, 300

Given that the number of observations is 24 (an even number).

The median is gotten by taking the 2 number at the middle 12th and 13th items which are 265 and 265.

Then the average of these numbers is calculated

Median = ½(265 + 265)

Median = ½(265 * 2)

Median = 265

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

3x + 4y -25 =0 represents the amount of change sianna should recieve