4. On a 300-mile auto trip, Lisa covered 52 mph for the first 100 miles, 65 mph for the second 100 miles, and 58 mph for the last 100 miles. The average speed covered by Lisa for the trip equals to a....


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4. On a 300-mile auto trip, Lisa covered 52 mph for the first 100 miles, 65 mph for the<br>second 100 miles, and 58 mph for the last 100 miles. The average speed covered by<br>Lisa for the trip equals to<br>a. 59<br>b. 58.09<br>с. 58.33<br>d. 57.85<br>5. A small country bought oil from three different sources in one week, as shown in the<br>following table.<br>Source<br>Меxico<br>Price per Barrel (S)<br>Barrels Purchased.<br>1000<br>200<br>51<br>Kuwait<br>64<br>Spot Market<br>The mean price per barrel for all 1300 barrels of oil purchased in that week equals to<br>а. 61.667<br>b. 433.33<br>100<br>70<br>c. 54.46<br>d. 64<br>

Extracted text: 4. On a 300-mile auto trip, Lisa covered 52 mph for the first 100 miles, 65 mph for the second 100 miles, and 58 mph for the last 100 miles. The average speed covered by Lisa for the trip equals to a. 59 b. 58.09 с. 58.33 d. 57.85 5. A small country bought oil from three different sources in one week, as shown in the following table. Source Меxico Price per Barrel (S) Barrels Purchased. 1000 200 51 Kuwait 64 Spot Market The mean price per barrel for all 1300 barrels of oil purchased in that week equals to а. 61.667 b. 433.33 100 70 c. 54.46 d. 64
1. Consider the following two data sets.<br>Data Set I<br>12<br>25<br>37<br>41<br>Data Set II<br>19<br>32<br>44<br>15<br>48<br>Notice that each value of the second data set is obtained by adding 7 to the<br>corresponding value of the first data set. The mean and standard deviation for Data<br>Set I is 24.6 and 14.6389 respectively. Then the mean and variance for Data Set II is<br>equal to<br>a. Mean = 24.6 and Variance = 214.3<br>b. Mean = 31.6 and Variance = 14.6389<br>c. Mean = 31.6 and Variance = 214.3<br>d. Mean = 24.6 and Variance = 14.6389<br>2. If the coefficient of variation for prices is 5.483% and the variance price Rs? 0.7225<br>then mean is equal to<br>a. 1.3177<br>b. 15.5<br>c. 7.5889<br>d. 0.5483<br>3. Through visualization not mathematically determine the location of mean, median<br>and mode classes from the given Histogram.<br>NUMBER OF PACKETS FOR POSTAGE<br>HAVING A CERTAIN WEIGHT<br>10<br>9.9<br>10.1 10.2 10.a 104 10.5 10.6<br>WEIGHT OF PACKETS/KG<br>Page 1 of 3<br>a. Mean = 10.3, median = 10.1,mode = 10.1<br>b. Mean = 10.01, median = 10.2, mode = 10.1<br>c. Mean = 10.0, median = 10.2, mode = 10.1<br>d. Mean = 10.3, median = 10.2, mode = 10.2<br>

Extracted text: 1. Consider the following two data sets. Data Set I 12 25 37 41 Data Set II 19 32 44 15 48 Notice that each value of the second data set is obtained by adding 7 to the corresponding value of the first data set. The mean and standard deviation for Data Set I is 24.6 and 14.6389 respectively. Then the mean and variance for Data Set II is equal to a. Mean = 24.6 and Variance = 214.3 b. Mean = 31.6 and Variance = 14.6389 c. Mean = 31.6 and Variance = 214.3 d. Mean = 24.6 and Variance = 14.6389 2. If the coefficient of variation for prices is 5.483% and the variance price Rs? 0.7225 then mean is equal to a. 1.3177 b. 15.5 c. 7.5889 d. 0.5483 3. Through visualization not mathematically determine the location of mean, median and mode classes from the given Histogram. NUMBER OF PACKETS FOR POSTAGE HAVING A CERTAIN WEIGHT 10 9.9 10.1 10.2 10.a 104 10.5 10.6 WEIGHT OF PACKETS/KG Page 1 of 3 a. Mean = 10.3, median = 10.1,mode = 10.1 b. Mean = 10.01, median = 10.2, mode = 10.1 c. Mean = 10.0, median = 10.2, mode = 10.1 d. Mean = 10.3, median = 10.2, mode = 10.2
Jun 02, 2022
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