Answer:
Step-by-step explanation:
We have
Put the coordinates of the given points and check the inequality:
For (0; 5) → x = 0; y = 5
For (2; 4) → x = 2, y = 4
For (1; 3) → x = 1, y = 3
<span>The
content of any course depends on where you take it--- even two courses
with the title "real analysis" at different schools can cover different
material (or the same material, but at different levels of depth).
But yeah, generally speaking, "real analysis" and "advanced calculus"
are synonyms. Schools never offer courses with *both* names, and
whichever one they do offer, it is probably a class that covers the
subject matter of calculus, but in a way that emphasizes the logical
structure of the material (in particular, precise definitions and
proofs) over just doing calculation.
My impression is that "advanced calculus" is an "older" name for this
topic, and that "real analysis" is a somewhat "newer" name for the same
topic. At least, most textbooks currently written in this area seem to
have titles with "real analysis" in them, and titles including the
phrase "advanced calculus" are less common. (There are a number of
popular books with "advanced calculus" in the title, but all of the ones
I've seen or used are reprints/updates of books originally written
decades ago.)
There have been similar shifts in other course names. What is mostly
called "complex analysis" now in course titles and textbooks, used to be
called "function theory" (sometimes "analytic function theory" or
"complex function theory"), or "complex variables". You still see some
courses and textbooks with "variables" in the title, but like "advanced
calculus", it seems to be on the way out, and not on the way in. The
trend seems to be toward "complex analysis." hope it helps
</span>
Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
Answer:
$29000 with a margin of error of $5000
Step-by-step explanation:
We have that the midpoint between the given values is
(X1+X2) / 2 = ($34000+$24000)/2 = $29000
We have that the midpoint between the given values would be
(X2-X1)/2=($34000-$24000)/2=$10000/2=$5000
So I can write that approach as $29000 with a margin of error of $5000
Done
Find the GCF of 80 and 32.
I'd start by identifying possible integer factors of both 80 and 32:
80: {1,2,4,5,8,10,16,20, 40, 80}
32: {1, 2,4, 8, 16, 32}
Working backwards, we see that the first factor that is represented in both lists is 16. Is 80 evenly divisible by 16? Yes; the quotient is 5.
Is 32 evenly divisible by 16? Yes; the quotient is 2.
You could writet 80 + 32 as 16(5 + 2). This is a product equal to 112, just as 80 + 32 = 112.
