Apacket of crips weighs 32 grams to the nearest gram. a multipack of crisps contain 10 packets. work out the least and greatest weights of the multipack. you can ignore the weight of the multipack wrapper.

its to do with upper and lower bounds

The least and the greatest weights are 315 grams and 325 grams

Step-by-step explanation:

* Lets explain what is the upper and lower bounds

- The lower bound is the smallest value that would round up to the

  estimated value.

- The upper bound is the smallest value that would round up to the next

 estimated value.

- Ex: a mass of 70 kg, rounded to the nearest 10 kg, has a lower

 bound of 65 kg, because 65 kg is the smallest mass that rounds to

 70 kg. The upper bound is 75 kg, because 75 kg is the smallest mass

 that  would round up to 80 kg, then 65 ≤ weight < 75

- So to understand how to find them divide the nearest value by 2

 and then subtract it and add it to the approximated value

* Lets solve the problem

- A packet of crisps weighs 32 grams to the nearest gram

- The nearest value is 1 gram

∴ 1 ÷ 2 = 0.5

- To find the lower bound subtract 0.5 from the approximated value

∵ The approximated value is 32

∴ The lower bound = 32 - 0.5 = 31.5 grams

- To find the upper bound add 0.5 from the approximated value

∵ The approximated value is 32

∴ The upper bound = 32 + 0.5 = 32.5 grams

∴ 31.5 ≤ weight of one packet < 32.5

∵ A multipack of crisps contain 10 packets

- To find the least and greatest weights of the multipack multiply the

  the lower bound and the upper bound by 10

∵ The least value of one packet is 31.5

∴ The least weight of the mulipack = 31.5 × 10 = 315 grams

∵ The greatest value of one packet is 32.5

∴ The greatest weight of the mulipack = 32.5 × 10 = 325 grams

∴ 315 ≤ weight of multipack < 325

* The least and the greatest weights are 315 grams and 325 grams

the image of (6, 9) under a dilation is (4, 6). the scale factor is

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.