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2 years ago

Find the product -a2 b2 c2 (a+b-c)

1 answer:

7 0

Use the distributive property:
$a(c+b)=ab+ac$
And
$a^n\cdot a^m=a^{n+m}\\\\a=a^1$

$-a^2b^2c^2(a+b-c)=-a^2ab^2c^2-a^2b^2bc^2+a^2b^2c^2c=-a^3b^2c^2-a^2b^3c^2+a^2b^2c^3$

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A bee flies at 10 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for 18 minutes, and the

ololo11 [35]

Step by Step answer

Yes I believe that is correct

4 0

1 year ago

At her birthday party, Ms. Willow would not give her age directly. She said, 'If you add the year of my birth to this year, s

Nimfa-mama [501]

Answer:

69 years

Step-by-step explanation:

Given that:

Let year of birth = x

This year = 2020

Year of 25th birthday = x + 25

Year of 55th birthday = x + 55

Present age = 2020 - x

Hence,

x + 2020 - (x + 25 + x + 55) + (2020 - x) = 58

x + 2020 - (2x + 80) + (2020 - x) = 58

x + 2020 - 2x - 80 + 2020 - x = 58

-2x + 4040 - 80 = 58

-2x + 3960 = 58

-2x = 58 - 3960

-2x = - 3902

x = 1951

Birth year = 1951

Hence, age =

2020 - 1951 = 69 years

5 0

2 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

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