What are the zeros of the function defined by x^2+9+14

WIL L GIVE BRAINLIST NEED QUICK

Ivanshal [37]

Answer:

circumference of pool=81.67

Step-by-step explanation:

7 0

3 years ago

use coordinate notation to write the rule that maps each preimage to its image. The identify the transformation and confirm that

Ymorist [56]

The rigid transformations used for each figure:

Figure 5 - Reflection around x and y axes: (x, y) → (- x, - y)
Figure 6 - Horizontal and vertical translations: (x, y) → (x + 1, y - 2)

<h3>What transformation rules do create the resulting images?</h3>

In this question we must determine what kind of rigid transformations generates each image. Rigid transformations are transformations applied on geometric loci such that Euclidean distance is conserved. Now we proceed to determine the transformation rule for each case:

Figure 5 - Reflection around the x-axis followed by reflection around the y-axis.

(x, y) → (- x, - y)

Figure 6 - Translation one unit in the +x direction and two units in the -y direction.

(x, y) → (x + 1, y - 2)

To learn more on transformation rules: brainly.com/question/9201867

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8 0

2 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

(2tens1one) times 10

Elodia [21]

Answer:

210

Step-by-step explanation:

10+10+1=21

21 x 10=210

6 0

3 years ago

Which of the following ratios can from a proportion with 5/6 ? Show your work!

Ludmilka [50]

Answer:

Step-by-step explanation:

5/6 = 14*5/14*65/6 = 5/6

7 0

2 years ago