Find the percent of discount. Round to nearest tenth.

How many times greater is the value of the fidget 2 in 234567 than the value of the digit 2 in 765432

Scilla [17]

Answer:

The value of 2 in the first digit is 100,000 times greater than the value of 2 in the second digit

Step-by-step explanation:

Firstly, we need to know the value of the 2 in 234,567 and that is 200,000

The value of the 2 in 765,432 is simply 2

So the number of times the value of the first 2 is greater than the value of the second 2 is;

200,000/2 = 100,000

So the value of 2 in the first digit is 100,000 times greater than the value of 2 in the second digit

6 0

3 years ago

A student says that if the hypothesis is false, an argument cannot be valid. How do you respond ?

Anton [14]

If a student tells me that an argument that has a false hypothesis cannot be valid, I would reply that we need to look carefully at the meaning of validity in the context of logic. In everyday speech, we tend to use “valid” to mean the same thing as “true” or “accurate.” In logic, this is not the way the term is used.

8 0

3 years ago

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If I cut a cake into equal parts and one slice = 3.333, where would I find the 0.001? (total of slices = 10)

ivolga24 [154]

Step-by-step explanation:

That is so because 1/3 is *not exactly equal to* 0.333333

0.333333 is an approximation of 1/3 which is correct to 6 decimal places.

So 0.333333 x 3 = 0.999999 which is approximately equal to 1.

If you want to be more accurate, you could write the fraction 1/3 everywhere and not write it in floating point form.

1/3 in floating point format will be 0.333333333…..and so on with infinite decimal places. This is because when you try to divide 1 by 3, you get a nonterminating recurring number after decimal point.

Its usually represented by a bar or dot on top of the recurring part

Ex. 1/3 = 0.3bar

(Sorry I can't get the bar on 3 by typing from keyboard)

But let's just say 0.3 bar is 0.333333…and so on to infinity.

When you multiply 0.3bar x 3 you get 0.9bar

= 0.999999….and so on to infinity.

These decimal point figures just a very great approximations of the fraction 1/3.

Hope this helps. Feel free to reply if you still have doubts.

8 0

3 years ago

In Sweden, 10 people are struck by lightning every year, on average. Answer thefollowing questions about a given year in Sweden

emmainna [20.7K]

Answer:

a) 19.85% probability that a total of two people are struck by lightning during first four months of the year.

b) 22.68% probability that the year has 5 good and 7 bad months

Step-by-step explanation:

We are going to use the Poisson distribution and the binomial distribition to solve this question.

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 time interval.

Binomial 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.

a.Find the probability that a total of two people are struck by lightning during first four months of the year.

10 people during a year(12 months).

In 4 months, the mean is $\mu = \frac{10*4}{12} = 3.33$

This is P(X = 2).

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

$P(X = 2) = \frac{e^{-3.33}*(3.33)^{2}}{(2)!} = 0.1985$

19.85% probability that a total of two people are struck by lightning during first four months of the year.

b.Say that a month is good is no one is struck by lightning, and bad otherwise. Find the probability that the year has 5 good and 7 bad months.

Probability that a month is good.

P(X = 0), Poisson

The mean is $\mu = \frac{10*1}{12} = 0.8333$

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

$P(X = 0) = \frac{e^{-0.8333}*(0.8333)^{0}}{(0)!} = 0.4346$

Find the probability that the year has 5 good and 7 bad months.

Now we use the binomial distribution, we want P(X = 5) when n = 12, p = 0.4346. So

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

$P(X = 5) = C_{12,5}.(0.4346)^{5}.(0.5654)^{7} = 0.2268$

22.68% probability that the year has 5 good and 7 bad months

3 0

3 years ago

Which is the same as 98.52 ÷ 10?

alexgriva [62]

Answer:

9.852 x 10 is the same outcome of 98.52/10. Can I have the brainliest answer?

Hope this helped!!!! :) <3 have a good day ma dude

7 0

3 years ago

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