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3 years ago

Round 68,413 to the nearest thousand

Mathematics

1 answer:

meriva3 years ago

5 0

8 is in the thousands place.

Look to the number in the hundreds (right next to it). It is a 4. Because 4 is less than 5, round down.

68413 rounded down in the thousands place becomes 68000

68000 is your answer

hope this helps

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Solve for w. <br>6w=-12 <br>Simplify your answer as much as possible.

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Answer:

-2

Step-by-step explanation:

Rearrange for w

-12/6 = -2

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Graph represents the function f (x) = StartFraction 5 minus 5 x squared Over x squared EndFraction?

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Answer:

The following are the attachment to this question.

Step-by-step explanation:

Given:

$\bold{ f (x) = (\frac{5- 5x^2}{x^2})}\\$

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$\bold{ f (x) = (\frac{-5 (x^2-1)}{x^2})}\\\\\bold{ f (x) = (\frac{-5(x+1)(x-1)}{x^2})}\\$

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Answer:

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Step-by-step explanation:

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3 years ago

What fraction of two pounds is twenty pence?

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Hey there

High light all of your key terms of this problems (fraction, of two pounds, and twenty pence)

Answer: We have $200$ pence of the $2$ $Lbs$ , which is also $20$ pence in the $\frac{1}{10} = 10%$

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4 years ago

Cassandra's Cogs advertises on its website that 90% of customer orders are received within three working days. They performed an

11111nata11111 [884]

Using the Central Limit Theorem, we have that:

a) Yes, it can be used, as np > 10 and n(1 - p) > 10.

b) There is a 0.0475 = 4.75% probability that the proportion in the random sample of 100 orders is the same as the proportion found in the audit sample or less.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

For the sample, we have that:

np = 85 > 10.

n(1 - p) = 15 > 10.

Hence the normal approximation can be used.

As for part B, if we have p = 0.9 and n = 100, the mean and the standard error are given by:

$\mu = p = 0.9$ .

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.9(0.1)}{100}} = 0.03$

The probability of a sample proportion of 85% of less is the <u>p-value of Z when X = 0.85</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.85 - 0.9}{0.03}$

Z = -1.67

Z = -1.67 has a p-value of 0.0475.

There is a 0.0475 = 4.75% probability that the proportion in the random sample of 100 orders is the same as the proportion found in the audit sample or less.

More can be learned about the Central Limit Theorem at brainly.com/question/24663213

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