What is 98,464 rounded to the nearest ten thousands ?

What is the width of a rectangle with length 14 cm and area 161 cm to the 2nd power?

WINSTONCH [101]

A=l*w
161=14w

161/14=w

11.5cm is the width

7 0

2 years ago

Two urns contain white balls and yellow balls. The first urn contains 2 white balls and 7 yellow balls and the second urn contai

Romashka [77]

-- The first urn has 9 balls in it all together, and 2 of them are white.
If you don't peek, then the prob of pulling out a white ball is 2/9 .

-- The second urn has 13 balls in it all together, and 3 of them are white.
If you don't peek, then the prob of pulling out a white ball is 3/13 .

-- The probability of being successful BOTH times is

(2/9) x (3/13) = ( 6/117 ) = about 0.0513 or 5.13% (rounded)

4 0

3 years ago

Read 2 more answers

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .