A particular salad contains 4 units of vitamin A, 5 units of vitamin B complex, and 2 mg of fat per serving. A nutritious soup contains 6 units of vitamin A, 2 units of vitamin B complex, and 3 mg of fat per serving. If a lunch consisting of these two foods is to have at least 10 units of vitamin A and at least 10 units of vitamin B complex, how many servings of each should be used to minimize the total number of milligrams of fat

2 servings of salad and 1 serving of soup

Step-by-step explanation:

In the given scenario the aim is to minimise the fat content of the food combination.

Fat content of soup is 3mg while fat content of salad is 2 mg.

Using Soup as 0 and Salad as 2 will not give the required vitamin content

The logical step will be to keep servings of soup to the minimum.

Let's see some combinations of salad and soup. Keeping serving of soup to the minimum of 1

1. 1 serving of salad and one serving of soup will contain 10 mg of vitamin A, 7 mg of vitamin B complex, and 3 mg of fat.

This will not work because amount of vitamin B complex is not up to 10 mg

2. 2 servings of salad and 1 serving of soup. Will contain 14 mg of vitamin A, 12 mg of vitamin B, and 7 mg of fat

This is the best option as we have amount of vitamin A and vitamin B complex in adequate quantity.

Also fat is lowest in this combination because soup the food with highest fat content is at minimum amount of one serving

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.