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3 years ago

What is the square root of 104 rounded to the nearest hundredth

2 answers:

6 0

Square root of 104 = 10.19803.....
Rounded to the nearest hundredth is 10.20 because the 8 rounds up the 9 and the 9 rounds 1 up to 2.

10.20

agasfer [191]3 years ago

4 0

The answer I got was 10.21.

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P(x)=x^6-11x^5+30x^4<br><br> Number 15

MA_775_DIABLO [31]

Answer:

what do we need to do with number 15

Step-by-step explanation:

8 0

3 years ago

Round each decimal number to the nearest tenth. Then, select the best estimate for the subtraction equation 3.21 − 1.19 = ___. h

Ymorist [56]

Answer:

the answer after rounding is 2

8 0

2 years ago

I need help solving this!

blagie [28]

Answer:

miles hybrid car went = 9.80 gal × 54.1 miles/gal

= 530.18 miles

132km × 0.621 = 81.972 miles

530.18 miles = 9.80 gal

1 mile = 9.80/530.18

= 0.018 gal

81.972 miles = 0.018 × 81.972

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3 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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