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3 years ago

Simplify 5^6 5^2 O A. 54 B. 254 C. 53 D. 54

Mathematics

2 answers:

Sholpan [36]3 years ago

5 0

Answer:

D. 54 That's the right answersl

igomit [66]3 years ago

3 0

Answer:

Step-by-step explanation:

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I need the answer no explanation

cricket20 [7]

Answer:

the answer is option D because it cant be division or multiplication and minus does not work

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3 years ago

Read 2 more answers

(p^2-7pq-q^2)+(-3p^2-2pq+7q^2)

Alex Ar [27]

Answer:

-2p² - 9pq + 6q²

Step-by-step explanation:

Add the like terms

p² + - 3p² = -2p²

-7pq + - 2pq = -9pq

-q² + 7q² = 6q²

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3 years ago

Graph the equation 5x+2y=10

7nadin3 [17]

Desmos graphing calculator is a very helpful tool!

5x+2y=10: Displayed in the image below

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3 years ago

A buoy is 8 km N 60° E of a boat as shown. How far east of the boat is the buoy?

Bumek [7]

Answer:13.86km

Step-by-step explanation: see attachment

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3 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)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 error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

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

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

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

Part a

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 error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

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

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

4 0

3 years ago