The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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0.00067
You always want the period to the right of the first number meaning you’ll start from there counting backwards because it is negative.
Answer:
Please find attached the required graph for the temperature time relationship
Step-by-step explanation:
The parameters given are arranged in a tabular form as follows;
At
Time Temperature
6 a. m. 56° F
7 a. m. 59° F
8 a. m. 62° F
9 a. m. 66° F
10 a. m. 70° F
11 a. m. 74° F
12 p. m. 74°F
1 p. m. 74°F
2 p. m. 74°F
3 p. m. 74°F
4 p. m. 74°F
5 p. m. 74°F
6 p. m. 74°F
7 p. m. 71°F
8 p. m. 68°F
9 p. m. 65°F
10 p. m. 62°F
11 p. m. 61.5°F
12 a. m. 61°F
With the above table as the data points, the graph can be drawn
From Excel, The Mean temperature is approximately 68.1°F
The standard deviation is 6.079 °F
Answer:
The slope is -3
Step-by-step explanation:
16:6 is an equivalent ratio
