What is the average rate of change of the function f(x)=5(2)^x from x=1 to x=5

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the

insens350 [35]

Answer:

So, the volume V is

$V=\frac{14\pi}{3}$

Step-by-step explanation:

We have that:

$y=7x\\\\y=0\\\\x=1\\\\x=-4\\$

We have the formula:

$V=2\pi\int_a^b x(g(x)-f(x))\, dx\\\\g(x)>f(x).$

We calculate the volume V, we get

$V=2\pi\int_a^b x(g(x)-f(x))\, dx\\\\V=2\pi\int_0^1 x(7x-0)\, dx\\\\V=2\pi\int_0^1 7x^2\, dx\\\\V=2\pi\cdot 7\left[\frac{x^3}{3}\right]_0^1\\\\V=14\pi\left(\frac{1}{3}-\frac{0}{3}\right)\\\\V=\frac{14\pi}{3}$

So, the volume V is

$V=\frac{14\pi}{3}$

We use software to draw the graph.

6 0

3 years ago

What is the distance formula?

alina1380 [7]

Answer:

A

Step-by-step explanation:

4 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

Still in class please need this fast

ivanzaharov [21]

Answer:

12 km

Step-by-step explanation:

C is 5 km

4+3+5=12

pls brainliest

7 0

2 years ago

Read 2 more answers

Theres a picture, im giving brainliest

Thepotemich [5.8K]

Answer:

A maybe?

Step-by-step explanation:

3 0

2 years ago