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3 years ago

Find the surface area of a cube is 150 cm2. What is the length an edge of

Mathematics

2 answers:

Alex Ar [27]3 years ago

6 0

Answer:

Step-by-step explanation:

The formula is 6x^2

If you set it equal to 150

Svet_ta [14]3 years ago

6 0

Answer:

5 x 5 + 5 x 5 + 5 x 5 + 5 x 5 + 5 x 5 + 5 x 5 = 25 + 25 + 25 + 25 + 25 + 25 = 50 + 50 + 50 = 150

5 would be the missing side because it fits the formula

(A x A) + (A x A) + (A x A) + (A x A) + (A x A) + (A x A) = Surface area

Step-by-step explanation:

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Four friends share 3 fruit bars equally. what fraction of a fruit bar does each person get?

oee [108]

Answer:

3/4

Step-by-step explanation:

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3 years ago

PLEASE HELP!

spin [16.1K]

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2 years ago

Martha divides $94 amongst her friends. Leon gets twice as much money as K. Jill gets five more dollars than Leon. Isaac gets $1

vlada-n [284]

Let me see if I can solve this on my own.

So $94 is the first amount.

It got divided so, 94÷2 = $47

So, It's either $47 or this.

47 + 47 (or times 2) = 94 again. But gets five more dollars than Leon. So 94+5= 99 then -10 = 89 but 89 ÷ 2 = 44.5 but since money can't really be half. It's eithier $44 or $45 but in my opinion is $45.

So, these are the most reasonable answers. $47 or $45. You can pick the answer that you thinks makes more sense. This took a long time so I hope this helps!

4 0

3 years ago

Read 2 more answers

Help ill give brainlest pls

Murrr4er [49]

Answer:

Step-by-step explanation:

Label the coordinate plane with the numbers so you can see it, you can also use a coordinate plane from online for a reference

8 0

2 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

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

In which

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

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

2 years ago