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3 years ago

What number would complete the pattern 16,4,12,36,9,27,44,11

1 answer:

6 0

<span>This isn't very difficult to figure out, you just need to look at these numbers in pairs. So, 16 minus 4 is 12. Then the next pair of numbers, 36 minus 9 is 27. And to complete the pattern, you just have to look at the final pair: 44 minus 11 is 33. So the number that would complete the pattern is 33.Hope this helps. Let me know if you need additional help!</span>

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What's 9 + 10-4 + 78

Zarrin [17]

Answer:

Step-by-step explanation:

78+9+(10-4)=

78+9+6=

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3 years ago

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Use the figure below to find the value of x and PQ.

Scilla [17]

Answer:

x=14

PQ=46

Step-by-step explanation:

PT and TQ are equal, so we can set up the equation like this: PT=TQ

23=2x-5

calculate to find x

x=14

PQ is just PT+TQ so 23+23=46

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2 years ago

How to slove -10 -1 need help

yuradex [85]

Answer:

-10 - 1 = x

Step-by-step explanation:

If you take -10 and subtract one, it makes the number go down lower because it's negative (example, 12 - 2 is -14)

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3 years ago

Solve for .6u+7u-16=-81<br> Simplify your answer as much as possible.

Vesnalui [34]

Answer:

6u+7u-16=-81

We simplify the equation to the form, which is simple to understand

6u+7u-16=-81

We move all terms containing u to the left and all other terms to the right.

+6u+7u=-81+16

We simplify left and right side of the equation.

+13u=-65

We divide both sides of the equation by 13 to get u.

u=-5

Step-by-step explanation:

I think i did it correct

7 0

3 years ago

Read 2 more answers

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