A training field is formed by joining a rectangular and two semicircles. The rectangle is 95 m long and 74m wide. Find the area

sergeinik [125]

Answer:

The slope is 3/2.

Step-by-step explanation:

The slope of a line is the relationship between the change of the y-variable and the change of the x-variable. It can be denoted as m = (y2-y1)/(x2-x1), and in layman's terms, it is known as rise over run.

You always can obtain the slope of a line with two points. From this line, for example, I can tell that points (0, 20) and (20, 50) are on the line.

To find the slope of the line, we must first find out what x1, x2, y1, and y2 are. x1 and y1 are the x- and y- coordinates of the first point, and x2 and y2 are the x- and y-coordinates of the second point. With this, we now know that x1 = 0, y1 = 20, x2 = 20, and y2 = 50. We now subsitute this into the formula.

m = (y2-y1)/(x2-x1)

m = (50 - 20)/(20 - 0)

m = (30)/(20)

m = 3/2

Now we know that the slope of the line is 3/2.

7 0

3 years ago

Read 2 more answers

Helppppp meeeeeeeeeeeeeeeeee

kati45 [8]

Answer:

The length of the shortest piece is $1\textrm{ ft}$ .

The length of the medium-sized piece is $3\textrm{ ft}$ .

The length of the longest piece is $7\textrm{ ft}$ .

Step-by-step explanation:

The five-step method is GUESS method.

Given:

Length of copper wire, $L=11\textrm{ ft}$

Length of longest piece is 6 feet longer than the shortest one.

Length of the medium-sized piece is 3 times the shortest one.

Unknown:

Length of longest, medium-sized and shortest pieces of wire.

Let the length of shortest piece be $x$ feet.

Therefore, the length of longest piece is equal to $6+x$ .

And, length of the medium-sized piece is equal to $3x$ .

Evaluate:

Sum of lengths of longest, medium-sized and shortest pieces of wire is equal to the total length of wire. So,

$6+x+x+3x=11\\5x+6=11\\5x=11-6\\5x=5\\x=\frac{5}{5}=1$

Substitute and Solve:

Now, we have, $x=1$ .

Substitute 1 for $x$ and solve for each sizes of wire pieces. This gives,

Length of shortest piece = $x=1\textrm{ ft}$

Length of medium-sized piece = $3x=3(1)=3\textrm{ ft}$

Length of longest piece = $6+x=6+1=7\textrm{ ft}$

Therefore, the lengths of shortest, medium-sized, and longest pieces are $1\textrm{ ft}$ , $3\textrm{ ft}$ and $7\textrm{ ft}$ respectively.

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)

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