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Pani-rosa [81]
3 years ago
15

List the numbers that have 30 as a multiple

Mathematics
1 answer:
olga2289 [7]3 years ago
3 0
First we list the multiples of each number.
multiples of 5 are 10,15,20,25,30,35,40,...
multiples of 6 are 12,18,24,30,36,42,48,...
multiples of 15 are 30,45,60,75,90,...
now, when you look at the list of multiples, you can see that 30 is the smallest number that appears in each list.
therefore, the least common multiple of...... 5,6 and 15 is 30.
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An insurance company has 25,000 automobile policy holders. If the yearly claim of a policy holder is a random variable with mean
telo118 [61]

Answer:

P(T>8300000)=1-P(T

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Data given

n = 25000 represent the automobile policy holders

\mu= 320 represent the population mean

\sigma =540 represent the population standard deviation

Let T the variable that represent the total of interest on this case. We can assume that the random variable for an individual policy holder is given by:

X\sim N(\mu = 540, \sigma=540)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

First we need to find the distribution for the random variable T like this:

\bar X = \frac{\sum_{i=1}^n x_i}{n}

And the total T is given by:

T=\sum_{i=1}^n X_i =n \bar X

We can find the expected value, variance and deviation for this random variable like this:

E(T)= n E(\bar X) = n \mu = 25000*320=8000000

Var(T)= Var(n\bar X)= n^2 Var(\bar X) = n^2 \frac{\sigma^2}{n}=n \sigma^2 =25000*(540^2)=7290000000

Sd(T)=\sqrt{7290000000}=85381.497

And we are interested on this probability:

P(T>8300000)

And we can use the Z score formula given by:

Z=\frac{T-E(T)}{\sigma_T}

P(T>8300000)=1-P(T

6 0
3 years ago
211 times 9=<—round the answer.
Yuki888 [10]

Answer:

211 × 9 is 1899 so rounding to nearest thousands would be 2000

3 0
3 years ago
Express as a fraction in simplest form: 43.5%
valkas [14]

Answer:

43.5/100

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Is 54,000 great or less than 6 km
Kazeer [188]
Less than 6km.............................................
                                    
5 0
3 years ago
Read 2 more answers
1 poir
poizon [28]

9514 1404 393

Answer:

  $1790.99

Step-by-step explanation:

<u>Given</u>:

  $1625 is invested at an annual rate of 1.95% compounded quarterly for 5 years

<u>Find</u>:

  the ending balance

<u>Solution</u>:

The compound interest formula applies.

  FV = P(1 +r/n)^(nt) . . . Principal P at rate r for t years, compounded n per year

  FV = $1625(1 +0.0195/4)^(4·5) = $1625(1.004875^20) ≈ $1790.99

The account ending balance would be $1790.99.

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