Write an expression showing the sum of 8 and a number f.<br> Show Your Work

Work out 5 x 10^6 / 4 x 10^4<br> Give your answer as an ordinary number.

Angelina_Jolie [31]

Answer:

$1.25 \times {10}^{2} = 125$

Step-by-step explanation:

$\frac{5 \times {10}^{6} }{4 \times {10}^{4} } = \frac{5}{4} \times \frac{ {10}^{6} }{ {10}^{4} } = 1.25 \times {10}^{2} = 125$

5 0

3 years ago

The cost to purchase a song from website 1 is $0.30 per song and a $5 membership fee. To purchase a song from website 2 there is

pashok25 [27]

Answer:

I belive the answer is 20 songs.

Step-by-step explanation:

The equation you're looking for to solve this is

0.30x+5=0.20x+7

Then, you would solve for x in order to get 20.

5 0

3 years ago

If f(x) = 2x + 1, find f(1).

____ [38]

Answer:

3

Step-by-step explanation:

5 0

3 years ago

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample

fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$