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2 years ago

Factor the following by completing the square w^4-22w^2+9

Mathematics

1 answer:

algol [13]2 years ago

3 0

Answer:

(w² - 4w - 3) (w² + 4w - 3)

Step-by-step explanation:

w² + 4w³ - 3w² -4w³ - 16w² + 12w-3w² - 12w + 9

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The point (5, -2) is rotated 90° about the origin. Its image is

tigry1 [53]

rotated 90° about the origin
(x,y) = (-y, x)

so (5, -2) is rotated 90° about the origin = (2, 5)

answer
b.(2, 5)

7 0

2 years ago

Students took a typing test. The number of words per minute students could type are listed below. 40, 47, 25, 32, 48, 55, 33, 56

const2013 [10]

Answer:

Step-by-step explanation:

The median number is known as the middles number

The first step is to rearrange the set of numbers starting from the lowest

25, 32, 32, 32, 33, 40, 47, 47, 48, 55, 56

Hence the middle number is 50

8 0

2 years ago

11) It's perimeter

natta225 [31]

2.8 m = .0028 km
3.5/.0028 = 1259 revolutions

8 0

2 years ago

Please help me with 5

Fantom [35]

Bluegill
3/5 × 1100
660

catfish
70% × (1100-660)
70% × 440
308

so not catfish or blue gill
1100 - (660 + 308) = 132

6 0

2 years ago

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0

2 years ago