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insens350 [35]
4 years ago
9

Please help this is clevermilkshakes sister

Mathematics
1 answer:
mr_godi [17]4 years ago
7 0

Answer:

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Step-by-step explanation:

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Someone help now please
azamat

Answer

7.000,819

Explanation

Good Luck ^-^

7 0
3 years ago
Read 2 more answers
Find SP where MP=Rs875 and discount =Rs75
liraira [26]

Answer:

SP is Rs 800.

Step-by-step explanation:

MP = Rs 875

Discount = Rs 75

so

SP = MP - discount

= Rs 875 - Rs 75

= Rs 800

4 0
3 years ago
Wich circle has a radius that measures 10 units?
LenaWriter [7]

c, 10 G this is the answer i have to talk cause it made me have to.

5 0
4 years ago
How do you solve this
fiasKO [112]
(((x + y) / 3) + (1 / x)) / (5 + (15 / x))
The best way is to make it one fraction.
Multiply by ((3x/3x) / (3x/3x)) to remove the other fractions.
((x(x + y)) + 3(1)) / (5(3x) + 3(15))
(x^2 + xy + 3) / (15x + 45)
Then factor to simplify
(x^2 + xy + 3) / (3(x + 15))



4 0
3 years ago
You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not a
11111nata11111 [884]

Answer:

For the exponential distribution:

\mu = \frac{1}{\lambda}

\sigma^2 = \frac{1}{\lambda^2}

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

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

\mu_{\bar X} = \bar X

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}

Step-by-step explanation:

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".

For this case we have a large sample size n =60 >30

The exponential distribution is the probability distribution that describes the time between events in a Poisson process.

For the exponential distribution:

\mu = \frac{1}{\lambda}

\sigma^2 = \frac{1}{\lambda^2}

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

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

\mu_{\bar X} = \bar X

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}

5 0
4 years ago
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