All categories

2 years ago

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

Mathematics

1 answer:

barxatty [35]2 years ago

5 0

Answer:

a/b tends to an infinite value

Step-by-step explanation:

If If a is an arbitrary nonzero constant, and we are to look for a/b as b approaches zero, we can represent this statement using limits. The statement is expressed as:

$\lim_{b \to 0} \dfrac{a}{b}$

Substituting b = 0 into the function

$= \dfrac{a}{0} \\\\= \infty \\\\\lim_{b \to 0} \dfrac{a}{b} = \infty\\ \\\\$

Since the limits of a tends to infinity as b tends to zero hence we can conclude that If a is an arbitrary nonzero constant then a/b tends to infinity or is undefined as b approaches 0

You might be interested in

Which mathematical concepts were the result of the work of René Descartes? Check all that apply. theory of an Earth-centered uni

Tanzania [10]

The appropriate response is the third one. A Cartesian organize framework is an arrange framework that determines each point exceptionally in a plane by a couple of numerical directions, which are the marked separations to the point from two settled opposite coordinated lines, measured in a similar unit of length. Each reference line is known as an organize pivot or only hub of the framework, and the point where they meet is its birthplace, as a rule at requested combine (0, 0).

4 0

2 years ago

Read 2 more answers

Write an equation for the 4th degree polynomial graphed below. Use k if your leading coefficient is positive and −k if your lead

Afina-wow [57]

Ho ho ho, lets get this party started
ok so I'm just really excited to use this stuff that I just learned

so

multiplicites
if a root or zero has an even multilicity, the graph bounces on that root
if the root or zero has an odd multiplicty, the graph goes through that root

so
roots are
-1
2
4
multiplicty is how many times it repeats
2 has even multiplity
we just do 2 is odd and 1 is even so
for roots, r1 and r2, the facotrs would be
(x-r1)(x-r2)
so
(x-(-1))^1(x-2)^2(x-4)
(x+1)(x-2)^2(x-4)
this is a 4th degre equaton
normally, it is goig from top right to top left
it is upside down
theefor it has negative leading coefient

y=-k(x+1)(x-4)(x-2)^2

8 0

2 years ago

Read 2 more answers

Ari is designing a patio that will be 20 feet long and 16 feet wide. He makes a scale drawing of the patio using a scale of 1 in

lara31 [8.8K]

Wouldnt it be four? id its 1/4

6 0

2 years ago

Read 2 more answers

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago

Mrs. Kermit has 29 markers. She gives 12 markers to Mr. Cooke and divides the rest among 5 other teachers. If each of the teache

romanna [79]

Answer: 2 markers.

Step-by-step explanation:

1. You know that she gives 12 markers to Mr. Cooke, then the rest among is:

29 markers-12 markers=17 markers

2. She divides the rest among 5 other teachers and each one of them get the same number of markers. So, you need to divide 17 markers by 5 teachers as following:

17 markers/5=3.4 markers

3. But the number of marker each one teacher has can't be a decimal number, so this number must be: 3 markers for each teacher.

3. The number of markers that will be left over is:

17 markers-(5*3 markers)=2 markers

8 0

2 years ago