Guven that the <span>distance
d that a certain particle moves may be calculated from the expression
where a and b are constants; and t is the elapsed time.
Distance is a length and hence the dimension of distance is L.
Now,
and
also will have the dimension of L.
</span><span>Time has a dimension of T.
For
, let the dimension of
be
, then
For
, let the dimension of
be
, then
Therefore, the dimension of
is
while the dimension of
is
.
</span>
Answer:
<h2>33</h2>
Step-by-step explanation:
Given the average low temperature by month in Nashville is represented by the function f(x)=-1.4x² + 19x +1.7, where x is the month, the average rate of change is expressed as d[f(x)]/dx = 2(1.4x) + 19
d[f(x)]/dx = 2.8x + 19
Since the number of months between March and August is 5 months and x is in months, hence we will substitute x = 5 into the resulting function to get the average rate of change from March to August as shown;
d[f(x)]/dx at x = 5
= 2.8(5)+ 19
= 14 + 19
= 33
<em>Hence the average rate of change from March to August is 33</em>
To calculate this, the Hardy-Weinberg principle can be used:
p² + 2pq + q² = 1 and p + q = 1
where p and q are the frequencies of the alleles (p - dominant, q - recessive), and p², q² and 2pq are the frequencies of the genotypes.
a) Since 32 plants have rough seed (recessive genotype: q²) out of 100 plants in total, then
q² = 32/100 = 0.32
b) q = √q² = √0.32 = 0.56
c) Since p + q = 1, then
p = 1 - q = 1 - 0.56 = 0.44
d) 19 plants with rough seeds (recessive genotype: q²) in a population of 100 means that q² = 19/100 = 0.19
We need to calculate p (the allele frequency for smooth seeds).
We can find q because we know q²:
q = √q² = √0.19 = 0.44
Since p + q = 1, then
p = 1 - q = 1 - 0.4 = 0.56
Answer:
a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)
=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363
hence probability that it actually contains explosives given identified as containing explosives)
=(10/(4*106))*0.95/0.005002363=0.000475
b)
let probability of correctly identifying a bag without explosives be a
hence a =0.99999763 ~ 99.999763%
c)
No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,
we have
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Factor the leading coefficient
----------> the value of A is 
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
therefore
the answer is
the value of A is
