The diameter of the bottom of a lamp shade is twice the diameter of the top of the lamp shade. The top of the lamp shade has a 9
1 answer:
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Let
A------> <span>(5√2,2√3)
B------> </span><span>(√2,2√3)
we know that
</span>the abscissa<span> and the ordinate are respectively the first and second coordinate of a point in a coordinate system</span>
the abscissa is the coordinate x<span>
step 1
find the midpoint
ABx------> midpoint AB in the coordinate x
</span>ABy------> midpoint AB in the coordinate y
<span>
ABx=[5</span>√2+√2]/2------> 6√2/2-----> 3√2
ABy=[2√3+2√3]/2------> 4√3/2-----> 2√3
the midpoint AB is (3√2,2√3)
the answer is
the abscissa of the midpoint of the line segment is 3√2
see the attached figure
A------> <span>(5√2,2√3)
B------> </span><span>(√2,2√3)
we know that
</span>the abscissa<span> and the ordinate are respectively the first and second coordinate of a point in a coordinate system</span>
the abscissa is the coordinate x<span>
step 1
find the midpoint
ABx------> midpoint AB in the coordinate x
</span>ABy------> midpoint AB in the coordinate y
<span>
ABx=[5</span>√2+√2]/2------> 6√2/2-----> 3√2
ABy=[2√3+2√3]/2------> 4√3/2-----> 2√3
the midpoint AB is (3√2,2√3)
the answer is
the abscissa of the midpoint of the line segment is 3√2
see the attached figure
Answer:
There is a 24.3% probability that one of the calculators will be defective.
Step-by-step explanation:
For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability of a defective calculator is 10 percent.
This means that
If 3 calculators are selected at random, what is the probability that one of the calculators will be defective
This is P(X = 1) when n = 3. So
There is a 24.3% probability that one of the calculators will be defective.