Simplify each expression 7x +5(x +3) +4x -x -2

An item is regularly priced at 85$. It is on sale for 80% off the regular price. How much is discounted from the regular price?

laila [671]

$68 is discounted from the regular price.

To find this, you first divide 80% by 100 to get 0.8 (the decimal form of the percentage). Then, simply multiply 0.8 by 85 (or vise verse) to get $68. Since the question is asking for how much is discounted from the regular price, the final answer would be $68.

I hope this helps!

4 0

2 years ago

Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an

Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

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.

We have:

Mean $\mu$ , standard deviation $\sigma$ .

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

8 0

2 years ago

What is the value of the expression below? 64^4/6

bonufazy [111]

Answer:

64^(4/6) = 16

Step-by-step explanation:

8 0

3 years ago

Life rating in Greece. Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sa

elena-s [515]

Answer:

a

The population parameter of interest is the true proportion of Greek who are suffering

While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%

b

The condition is met

c

The 95% confidence interval is $0.223 < p < 0.277$

d

If the confidence level is increased which will in turn reduce the level of significance but increase the critical value( $Z_{\frac{\alpha }{2} }$ ) and this will increase the margin of error( deduced from the formula for margin of error i.e $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

e

Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower

Step-by-step explanation:

From the question we are told that

The sample size is n = 1000

The population proportion is $\r p = 0.25$

Considering question a

The population parameter of interest is the true proportion of Greek who are suffering

While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%

Considering question b

The condition for constructing a confidence interval is

$n * \r p > 5\ and \ n(1 - \r p ) >5$

So

$1000 * 0.25 > 5 \ and \ 1000 * (1-0.25 ) > 5$

$250 > 5 \ and \ 750> 5$

Hence the condition is met

Considering question c

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_\frac{ \alpha }{2} * \sqrt{ \frac{\r p (1 - \r p ) }{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{ 0.25 (1 - 0.25 ) }{ 1000} }$

$E = 0.027$

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.25 - 0.027 < p < 0.25 + 0.027$

substituting values

$0.223 < p < 0.277$

considering d

If the confidence level is increased which will in turn reduce the level of significance but increase the critical value( $Z_{\frac{\alpha }{2} }$ ) and this will increase the margin of error( deduced from the formula for margin of error i.e $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

considering e

Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower

5 0

3 years ago

2 + 3 × 16 − 2 × 21 − 3 = 5 14 35 40

Ganezh [65]

Answer: 5

Step-by-step explanation:

Since, According to the BODMAS, when we solve an expression,

We follow the following sequence,

1. Bracket/ parenthesis

2. Of

3. Division

4.Multiplication