Answer:
Part a) The distance on a map between Joseph's house and the airport is 2.53 inches
Part b) The distance on a map between the airport and the restaurant is 1.68 inches and the total distance on a map between Joseph's house and the restaurant is 4.21 inches
Step-by-step explanation:
Part a) The actual distance between Joseph's house and the airport is 24 miles. How far apart are Joseph's house and the airport on the map?
we know that
The scale of a map is 1 inch : 9.5 miles
so
using proportion
Find the distance on a map if the actual distance between Joseph's house and the airport is 24 miles
Let
x-----> the distance on a map
1/9.5=x/24
x=24/9.5=2.53 inches
Part b) Joseph traveled from his house to the airport. He then traveled another 16 miles past the airport to a restaurant. How many inches on the map represent this distance?
we know that
The scale of a map is 1 inch : 9.5 miles
so
using proportion
Find the distance on a map if the actual distance between airport to the restaurant is 16 miles
Let
x-----> the distance on a map
1/9.5=x/16
x=16/9.5=1.68 inches
The total distance on a map between Joseph's house and the restaurant is equal to
2.53 inches+1.68 inches=4.21 inches
Answer:
Hello so.
65.4 x 2.3 = 150.42
Step-by-step explanation:
The Images I have attached show the work.
Answer:
I believe it is 49.425
Step-by-step explanation:
You just add all the numbers then divide it by how many numbers there are.
Answer:
200
Step-by-step explanation:
The linear model of this case takes the form:
y = a(x-b) + k
<span>The cost of having a package delivered has a base fee of $9.70
this is "k" >>>>> k=9.7 (fixed amount of fee)
THEN
</span>
<span>Every pound over 5 lbs cost an additional $0.46 per pound
that means: 0.46(x-5)
in other words, if the package weighs foe example 9 pounds, then 9-5=4, it will cost 0.46*4 for these 4 extra pounds
Finally we have the linear form of this: C = 0.46 (W - 5) + 9.7
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