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2 years ago

12

Jackson's kitten weighed 2 pounds 3ounces . A month later the kitten weighed 56 ounces. How much did the kitten gain in that mon

th

1 answer:

Novay_Z [31]2 years ago

5 0

Answer:

the kitten gained 21 ounces

Step-by-step explanation:

2 pounds = 32 ounces

so then i added 32 to 3 and got a total of 35 ounces

56-35=21

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Adam and Paul an achieve the sales target of the company in 30 days, Paul and Charles In 40 days while Charles and Adam In 60 da

Irina18 [472]

Answer:all three of them achieve the sales target together in 120 days.

Step-by-step explanation:

Given: Adam and Paul an achieve the sales target of the company in 30 days, Paul and Charles In 40 days while Charles and Adam In 60 days.

Then, the number of days taken by all of them to achieve the sales target together = $\dfrac1{30}+\dfrac{1}{40}+\dfrac{1}{60}$

$=\dfrac{1}{120}$ [LCM(30,40,60)=120]

Thus, all three of them achieve the sales target together in 120 days.

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2 years ago

Ryan opens a bank account with $50. Each month he deposits an amount that doubles his savings. Ryan gets $2 from the bank each m

irga5000 [103]

50(2+2) i think this is right

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2 years ago

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Use the graph to write an explicit function to represent the data and determine how much money Amy earned in week 5.

zalisa [80]

Answer:

3rd option

$f(n)=5(3)^{n-1}$

$f(5)=405$

Step-by-step explanation:

So we are given the following points:

(1,5)

(2,15)

(3,45)

(4,135)

This is a geometric sequence because there is a common ratio, 3. That is you can keep multiply 3 to a previous y-coordinate to get the next y-coordinate.

The formula for a geometric sequence is $f(n)=a_1 \cdot r^{n-1}$

where $a_1$ is the first term and r is the common ration.

So we have

$f(n)=5 \cdot (3)^{n-1}$ .

If you want to know the fifth term, just plug in 5:

$f(5)=5 \cdot (3)^{5-1}$

Simplifying:

$f(5)=5 \cdot 3^4$

$f(5)=5 \cdot 81$

$f(5)=405$

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2 years ago

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Alex17521 [72]

1777/10000. might simplify

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2 years ago

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Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records

madreJ [45]

Answer:

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

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".

Solution to the problem

For this case we have the following info related to the time to prepare a return

$\mu =90 , \sigma =14$

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And the standard deviation would be:

$\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2$

And the best answer would be

b. 2 minutes

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2 years ago