Rachel offers to go to the bakery to buy cookies and cupcakes

a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve a

Neporo4naja [7]

Answer:

Tangent at (1,2) has value $\frac{15}{4}$ . And (-2, 4) is the point of intersection of horizontal tangent.

Step-by-step explanation:

Given curve equation,

$y^2=x^3+3x^2\hfill (1)$

To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,

$2y\frac{dy}{dx}=x^3+3x^2$

$\implies \frac{dy}{dx}=\frac{3(x^2+2x)}{2y}$

At (1, 2), tangent line is,

$\frac{dy}{dx}|_{(1,2)}}=\frac{15}{4}$

To find point of intersection of horizontal tangent we have to do,

$\frac{dy}{dx}=0$

$\implies x(x+2)=0\implies x=0 or -2$

Thus,

At x=0, y=0

but snce,

$\lim_{y\to 0}\frac{dy}{dx}\to \infty$

at (0,0) there exist a vertical tangent. And,

At x=-2, y=4.

Thus (-2, 4) is the point of intersection of horizontal tangent.

6 0

2 years ago

What does the digit 7 represent in 170,280?

Sidana [21]

Answer:

that is the ten thousands place

Step-by-step explanation:

3 0

2 years ago

Clarissa's income puts her in the bottom tax bracket (10%) last year. During the same year, she earned $250 in dividends and $75

Art [367]

Answer:

75 in coupons.

250 in dividends.

profit of 600 - 425 = 175 from her stock investment.

her total income is 250 + 75 + 175 = 500.

if all of this is taxed at 10%, then her tax will be 500 * .1 = 50.

8 0

2 years ago

M a right angle. 2 lines form a right angle. Another line extends between the 2 lines to form 2 angles. The top angle is labeled

siniylev [52]

Answer:

i think it's congruent .

Step-by-step explanation:

8 0

2 years ago

Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0

2 years ago