Given the equation of the line (y-2)=3(x+1) what is the slope

What is two numbers whose difference is 25.6 There are plenty of numbers and digits has a difference of 25.6. we shall illustrate these values using the following.
1. 50.0 – 24.4 = 25.6
2. 100.1 – 74.5 = 25.6 
3. 150 – 124.4 = 25.6 
4. 78 – 52.4 = 25.6 
5. 90 – 64.4 = 25.6 
6. 88.5 -62.9 = 25.6

5 0

3 years ago

Write this decimal in a fraction in its simplest form.

sammy [17]

The simplified form of that decimal in fraction form is 3/20. Hope this helps.

5 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