Question

5 multiple choice 1 point jake has taken three tests in his math class. he now has a mean average score of 80 and a median grade of 81. if the range of his scores is 13 points, what are his three test scores? 73,81,86 72,81,85 72,81,87 74,81,87 75,81,88 6 multiple choice 1 point jasmine is in the checkout line of a grocery store. her basket contains 3 bags of chips at a price of $3.64 per bag, 8 jars of peanut butter at $2.54 per jar, 10 frozen pizzas at $4.76 per pizza, and 3 candy bars at $1.51 per bar. find the mode of the prices of all the items in her basket. $3.29 $3.64 $2.54 $3.47 $4.76

Explanation:

Question 5

Step1: Check median

For three scores, the median is the middle one. All options have 81 as the middle score, so median condition is satisfied for all.

Step2: Check mean

Mean = (sum of scores)/3 = 80, so sum should be \( 80\times3 = 240 \).

• For 73,81,86: Sum = \( 73 + 81 + 86 = 240 \). Check range: \( 86 - 73 = 13 \).
• For 72,81,85: Sum = \( 72 + 81 + 85 = 238

eq 240 \).

• For 72,81,87: Sum = \( 72 + 81 + 87 = 240 \), but range = \( 87 - 72 = 15

eq 13 \).

• For 74,81,87: Sum = \( 74 + 81 + 87 = 242

eq 240 \).

• For 75,81,88: Sum = \( 75 + 81 + 88 = 244

eq 240 \).

Step3: Check range

Only 73,81,86 has range 13 and sum 240.

Mode is the most frequently occurring value.

• Chips: 3 items at $3.64
• Peanut butter: 8 items at $2.54
• Frozen pizzas: 10 items at $4.76
• Candy bars: 3 items at $1.51

The price with the highest frequency is $4.76 (10 times), then $2.54 (8 times), then $3.64 and $1.51 (3 times each). Wait, wait, no: Wait, the mode is the price that appears most number of times. Wait, no: each item's price: chips are 3 items, all at $3.64? No, wait: 3 bags of chips, each at $3.64: so 3 instances of $3.64. 8 jars of peanut butter, each at $2.54: 8 instances of $2.54. 10 frozen pizzas, each at $4.76: 10 instances of $4.76. 3 candy bars, each at $1.51: 3 instances of $1.51. So the price that occurs most is $4.76 (10 times), then $2.54 (8), then $3.64 and $1.51 (3). Wait, but the options include $4.76. Wait, let's recheck:
Wait, the problem says "the mode of the prices of all the items". So each item's price: so for 3 chips: 3 times $3.64. 8 peanut butter: 8 times $2.54. 10 pizzas: 10 times $4.76. 3 candy bars: 3 times $1.51. So the frequency of $4.76 is 10, which is the highest. So mode is $4.76.
Wait, but let's check the options. The options are $3.29, $3.64, $2.54, $3.47, $4.76. So $4.76 is an option. Wait, but earlier miscalculation? Wait, no: 10 frozen pizzas, each at $4.76: so 10 items with price $4.76. 8 peanut butter at $2.54: 8 items. 3 chips at $3.64: 3. 3 candy at $1.51: 3. So the most frequent price is $4.76 (10 times), so mode is $4.76.

Answer:

73, 81, 86

Question 6