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1 year ago

I have the answer as 1728 but it’s saying it’s wrong

Mathematics

1 answer:

Digiron [165]1 year ago

8 0

Answer:

216 meter cubes

Step-by-step explanation:

The part where you got 1728 as the volume for the cube is right, but the question is asking how many 2x2x2 meter cubes can fit, meaning that you must find the volume of the meter cubes and divide the cube with a volume of 1728 to get the amount of cubes. In this case 2^3 is 8, so 1728 divided by 8 is 216, which means that 216 meter cubes with 2x2x2 dimensions can fit inside the larger cube.

Hope this helps! :D

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Which recursive definition could be used to generate the sequence {3, 6, 3, 6, 3,...}? A) a1 = 3 and an = an-1 + 3n

barxatty [35]

The answer is choice C

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You start off with the first term being a1 = 3

Then you add on 3 to get to 6 as the second term

The next thing to do is add on -3, which is the same as subtracting 3, to get back to 3 once more.

This pattern continues on forever. To ensure you add on either +3 or -3 to the previous term, we add on the general term 3*(-1)^n

Overall, the nth term is found by a(n) = a(n-1) + 3*(-1)^n, which is what choice C shows. I'm assuming your teacher meant to write 3(-1)^n instead of 3(-1)n

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3 years ago

Read 2 more answers

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

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