Ask question

Ask question

All categories

2 years ago

13

A clownfish is 2 meters below the surface of the ocean. How does its distance from the surface compare to the vertical distance

from the eyes of: the jumping dolphin? the flying seagull? the octopus?

1 answer:

mina [271]2 years ago

5 0

Answer: Um...is there supposed to be a link? I'll edit my answer if there is.

You might be interested in

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

Please helppp meee I need it done noww

bearhunter [10]

Answer:

7/2

Step-by-step explanation:

3 wholes is equal to 6 halves plus a half equals 7 halves

6 0

3 years ago

Zolox worked 38 hours last week. He had $88 deducted from his earnings for taxes. If he had $273 left after the deduction, how m

Virty [35]

273+88=361
361/38=9.5
Zolox made $9.50 per hour

6 0

3 years ago

Read 2 more answers

Mr. Autry bought a hat for $99.90. He also bought a pair of boots for $89.90. What was the total cost of the hat and boots inclu

Pepsi [2]

Answer:

$204.98

Step-by-step explanation:

Step 1: Add both item cost

$99.90 + 89.90 = 189.80$

Step 2: Find the sales tax of the result

$\frac{8}{100}\times 189.8 = 15.184$

Step 1: Convert 8% to fraction
$8\%= \frac{8}{100}$
Simplify
$\frac{8}{100} = \frac{2}{25}$
Step 2: Convert 189.80 to a fraction
$189.80 = \frac{189.8}{1}$
Step 3: Set it up
$\frac{2}{25}\times \frac{189.8}{1}$
Step 4: Cross divide
$189.80 \div 25 = 7.592\\2 \div 1 = 2$
Step 5: Multiply the result
$7.592 \times 2 = 15.184$

Step 3: Add the result from the total cost of both items

$189.80 + 15.184 = 204.98$

The total cost of the hats and boots including sales tax is 204.98 dollars

7 0

2 years ago

Solve for xxx.<br> Your answer must be simplified.<br> 15 < -5 x

katrin [286]

Answer:

x<−3

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers