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2 years ago

Math problem (easy) ………………………………………

Mathematics

2 answers:

wolverine [178]2 years ago

7 0

Answer:

1/6

Step-by-step explanation:

Decimal Form 0.16

yawa3891 [41]2 years ago

6 0

Answer:

7/24

Step-by-step explanation:

Hope this helps

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Determine whether the improper integral converges or diverges, and find the value of each that converges.

Brrunno [24]

Answer:

Converges at -1

Step-by-step explanation:

The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.

We will make use of integral by parts to determine:

let:

$u=x$ $dv=e^(2x)\cdot{dx}$

$du=dx$ $v=2\cdot{e^(2x)}$

$\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du$

$\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx$

$\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C$

We can therefore determine that if x tends to 0 the limit is -1

$\lim_{x \to \0} 2xe^2^x-2e^2^x=0-1=-1$

5 0

4 years ago

The graph shows the relationship between the radius and volume for many cones whose height is 6imches

borishaifa [10]

Answer:

the 3rd one

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Can someone help me with this plz??? I will mark u brainliest!!!

MrMuchimi

Answer:

a is 622

Step-by-step explanation:

5 0

3 years ago

An arithmetic sequence has this recursive formula: (a^1 =8, a^n= a^n-1 -6

eduard

Answer:

$a_n = 8 + (n - 1) (-6)$

Step-by-step explanation:

Given

$a_1 = 8$

Recursive: $a_{n} = a_{n-1} - 6$

Required

Determine the formula

Substitute 2 for n to determine $a_2$

$a_{2} = a_{2-1} - 6$

$a_{2} = a_{1} - 6$

Substitute $a_1 = 8$

$a_2 = 8 - 6$

$a_2 = 2$

Next is to determine the common difference, d;

$d = a_2 - a_1$

$d = 2 - 8$

$d = -6$

The nth term of an arithmetic sequence is calculated as

$a_n = a_1 + (n - 1)d$

Substitute $a_1 = 8$ and $d = -6$

$a_n = a_1 + (n - 1)d$

$a_n = 8 + (n - 1) (-6)$

Hence, the nth term of the sequence can be calculated using $a_n = 8 + (n - 1) (-6)$

3 0

3 years ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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 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".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

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

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

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

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

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

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago