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3 years ago

Not sure which one, please help or give me clues to solve it.

Mathematics

2 answers:

diamong [38]3 years ago

8 0

Its the last one i think

ella [17]3 years ago

6 0

It's the second one.
Hope this helps

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Find the sum of all multiples of 4 between 25 and 100

Mademuasel [1]

Answer:

Sum of all multiples of 4 between 25 and 100 is 1116.

Step-by-step explanation:

Multiples of 4 between 25 and 100 = 28,32,36,40..............,96

let's check the series whether it is an arithmetic progression

a1 = 28 , a2 = 32 , a3 = 36 , a4 = 40

Common difference (d)=a2-a1=32-28 = 4

=a3-a2=36-32 = 4

=a4-a3=40-36 = 4

Common difference exists then the series of numbers is an Arithmetic progression.

To find number of terms :

a(a1) = 28 , d = 4 , l( last term ) = 96

l = a + (n-1)d

96 = 28 + (n-1)4

96 - 28 = (n-1)4

68 = (n-1)4

68/4 = (n-1)

17 = n - 1

17 + 1 = n

18 = n

Sum of all terms:

a = 28 , d = 4 , n = 18 , l = 96

Sn = n/2 ( a + l )

S18 = 18 / 2 (28 + 96)

S18 = 9 (124)

S18 = 1116

Therefore,Sum of all multiples of 4 between 25 and 100 is 1116.

NOTE:

Here 100 is also a multiple of 4 but question is asked between 25 and 100 so 100 is not considered in multiples.

7 0

3 years ago

Executives from Six Flags, a well-known amusement park chain, had interest in constructing a Six Flags theme park in a location

natima [27]

Answer:

The mean of the sampling distribution for the sample proportion when taking samples of size 500 from this population is equal to 0.248.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$