A. x=the cost of one lunch. 60.25=5x+3(x-5.25)
b. 60.25=5x+3x-15.75 60.25+15.75=8x-15.75+15.75
76=8x 76/8=8x/8 9.50=x
A lunch is $9.50. If you are not sure, check my answers. Hope this helped.
To solve this question, you¨ll need to use a table of the standardized Normal distribution, or you could use a graphing calculator. The answer will be the cumulative property which will be a number with decimals. However, this represents the students who had a lower score then her, You´ll need to subtract this from 1 in order to get the opposite number, which will be the kids who scored higher. Be sure to multiply the result by 100 to get a percentage. Your answer should be 0.806 hope this helped!
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Circumference of circle = π × Diameter
= π × 18 cm
= 56.54867 or 56.54 to 1 dp