Answer:
970 cm
Step-by-step explanation:
Given that:
Size of brick :
25cm × 16cm × 10cm
Number of bricks = 4700
Dimension of wall = 20 m long × 20 cm wide
Let the height = h
Hence, dimension of wall (length and height)
2000 cm × h
Dimension of window = 2m × 1.5m = 200cm × 150cm
Number of windows = 2
Total area covered by windows = 2 * (L * B)
= 2(200 * 150) = 60,000cm
Brick = (25 * 16) * 4700 = 1, 880, 000 cm ( length and height)
Wall = 2000 * height - 60,000 = 1880000
2000height - 60,000 = 1880000
2000height = 1880000 + 60,000
2000height = 1940000
Height = 1940000 / 2000
Height of wall = 970 cm
Answer:
y=2x+5
Step-by-step explanation:
Hoped this helped
Answer:
It is A: 11 to 1 :)
It's true!
(edited my answer btw, you have to simply also welp)
The computed value must closely match the real value for a model to be considered valid. If the percentage of pleased or very satisfied students remains close to 75% after Mateo surveys additional students, Mateo's model is still viable. The model is faulty if the opposite is true.
<h3>How will mateo know whether his model is valid or not?</h3>
In general, a valid model is one whose estimated value is close to the real value. This kind of model is considered to be accurate. It must be somewhat near to the real value if it doesn't resemble the real value.
If the findings of the survey are sufficiently similar to one another, then the model may be considered valid.
P1 equals 75%, which is the real assessment of the number of happy pupils
P2 is 70 percent; this represents the second assessment of happy pupils
In conclusion, The estimated value of a model has to be somewhat close to the real value for the model to be considered valid. If the number of students who are either pleased or extremely satisfied remains close to 75 percent following Mateo's survey of more students, then Mateo's model is likely accurate. In any other scenario, the model cannot be trusted.
Read more about probability
brainly.com/question/795909
#SPJ1
The length would be 30 feet, and the width would be 10 feet.
