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.
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You might have made an error the first time you solved for x. I got x = -0.5.
When you have your log base 4, the way you cancel that out is by making 4 the base on both sides, so you get 4^(log4) to reduce to 1, and you're left with:
2x + 3 = 4^(1/2) ... Simplify
2x + 3 = 2
2x = -1
x = -1/2
If you plug that back in, everything checks out. Maybe double check your use of logarithm/exponent properties?
Answer:
D. 8
Step-by-step explanation:
The given diagram is a trapezium. We know that the consective sides of a trapezium are equal. so,
Putting the values of consecutive sides equal:
So, KI will be equal to LI
3x-7 = x+3
Putting the value of x in the equation of KI
3x-7
=3(5)-7
=15-7
=8
Hence, the correct answer is D. 8 ..
Answer:
$31.18
Step-by-step explanation:
14.50 x 2 = 29
29 x 0.075 = 2.175
29 + 2.175 = 31.175
Round up to 31.18
