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3 years ago

PLEASE HELP 25 POINTS!

Mathematics

2 answers:

yanalaym [24]3 years ago

7 0

Hey!

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Answer:

r = 2

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Solution:

So we know that s2 = 8 and s4 = 32.

We need to find s1!

32 / 8 = 4

8 / 4 = 2 (ratio is 2)

8 / 2 = 4

32 / 2 = 16

s1 = 4

s3 = 16

Each number is multiplied by 2.

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Hope This Helped! Good Luck!

liubo4ka [24]3 years ago

6 0

Answer:

The common ratio r = 2.

Step-by-step explanation:

Now s2 = a1r and s4 = a1r^3 where a1 = first term and r = common ratio so

s4 / s2 = a1r^3 / a1r = r^2 = 32/8

r^2 = 4

r = 2.

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An element with mass 490 grams decays by 28.6% per minute. How much of the element is remaining after 16 minutes, to the nearest

agasfer [191]

Answer:

2.2 grams

Step-by-step explanation:

We can use the compound decay formula shown below to solve this:

$F=P(1-r)^t$

Where

F is the final amount (what we want to find)

P is the present amount (490 grams)

r is the rate of decay, in decimal (28.6% = 28.6/100 = 0.286)

t is the time in minutes (16)

Substituting, we get:

$F=P(1-r)^t\\F=490(1-0.286)^{16}\\F=490(0.714)^{16}\\F=2.2355$

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3 years ago

Consider a parent population with mean 75 and a standard deviation 7. The population doesn’t appear to have extreme skewness or

Aleks04 [339]

Answer:

a) $\bar X \sim N(\mu=375, \sigma={\bar X}=\frac{7}{\sqrt{40}}=1.107)$

b) Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

c) $P(\bar X \leq 77)=P(Z$

d) See figure attached

Step-by-step explanation:

Previous concepts

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".

Let X the random variable of interest. We know from the problem that the distribution for the random variable X is given by:

$E(X) = 75$

$sd(X) = 7$

We take a sample of n=40 . That represent the sample size

Part a

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

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

$\bar X \sim N(\mu=375, \sigma={\bar X}=\frac{7}{\sqrt{40}}=1.107)$

Part b

Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

Part c

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$