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3 years ago

long beach calif has an elevation of -7 new orleans louisiana is 8 feet below sea level. which city has a lower elevation?

Mathematics

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

New Orleans Louisiana is lower

Step-by-step explanation:

California: 7 below sea level

Louisiana: 8 below sea level

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mamaluj [8]

Answer:

x = 18

y = 54

Step-by-step explanation:

A straight line equals 180 degrees.

Since Lines CD and AB are straight, we can deduce that the angle 3x = 2x + angle AN.

Because there is a right angle, we can subtract 90 -72 = 18.

18 + 2x = 3x

18 = x

A straight line equals 180 degrees. Once we figure out that x = 18, we can figure out y by plugging in x and then subtracting your answer from 180 degrees.

I hope this was helpful to you! If it was, please consider rating, pressing thanks, and giving my answer 'Brainliest.' Have a great day!

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3 years ago

Plz help me well mark brainliest!!

baherus [9]

Answer:

735 mm³

Step-by-step explanation:

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3 years ago

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Alex73 [517]

Answer:

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2 years ago

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Hlp!

zzz [600]

Answer:

Option B

x ≥ (-5)

Step-by-step explanation:

<h3><u>Given</u>;</h3>

-3(x + 4) ≥ x + 8

So,

-3(x + 4) ≥ x + 8

-3x – 12 ≥ x + 8

Add both sides 12 we get,

-3x – 12 + 12 ≥ x + 8 + 12

-3x ≥ x + 20

Similarly, subtract x from both sides we get,

-3x – x ≥ x – x + 20

-4x ≥ 20

Then, divide both sides by (-4) we get,

-4x/(-4) ≥ 20/(-4)

x ≥ -5

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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.

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3 years ago