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3 years ago

Ill give brainlist and 5 stars!

Mathematics

1 answer:

Kamila [148]3 years ago

6 0

Answer:

£24,301.55

Step-by-step explanation:

26705 decrease 9% =

26705 × (1 - 9%) = 26705 × (1 - 0.09) = 24301.55

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The elevation E, in meters, above sea level at which the boiling point of a certain liquid ist degrees Celsius is given by the f

sweet [91]

Answer:

Given E(t)=1100(100-t)+580(100-t)^2

Put t = 99.5, we get

E(99.5)=1100(100-99.5)+580(100-99.5)^2

E(99.5)=1100(0.5)+580(0.5)^2

E(99.5)=1100(0.5)+580(0.25)

E(99.5)=550+145

E(99.5)=695m

Step-by-step explanation:

7 0

3 years ago

If an object is moving at a speed of 6 meters per second for a time of 5 seconds, how far did the object travel? A. 35 meters B.

Cerrena [4.2K]

Answer: D 30 meters

Step-by-step explanation:

5 multiplied by 6 is 30

3 0

3 years ago

6x-11+2x=7+8x-17<br><br> A: Always true<br> B: Sometimes true<br> C: Never true

anastassius [24]

option C is the correct answer friend

7 0

3 years ago

A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa

UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so $\mu = \frac{14}{60} = 0.2333$

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081$

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0

3 years ago

a dog is going to be tied to a stake placed in the yard. the owners want the dog to have the largest space available in their tr

s2008m [1.1K]

Answer:

In the corner, so the dog has all the room of the triangular yard to roam and play.

6 0

3 years ago