A roll of fabric was 120 inches long. A customer bought 2 feet of the fabric.

Please help me. I really appreciate it if you answered. Thank you! <3

exis [7]

Answer:

$3100 is invested at 9%

$4900 is invested at 11%

Step-by-step explanation:

Let's take "x" be the amount invested at 9%.

(x + 1800) is invested in another account at 11%.

The interest amount earned by the two accounts is $818.

Here we can use the simple interest formula and find the amount invested in each account.

Simple interest (I) = , where P- is the principal , N is the number of years and R is the interest rate.

Simple interest =

0.09x + 0.11(x+1800) = 818

Now we have to simplify and find the value of x .

Use the distributive property and simplify the second term.

0.09x + 0.11x + 198 = 818

0.2x + 198 = 818

0.2x =818 - 198

0.2x = 620

x = 620/0.2

x = 3100.

So $3100 is invested at 9%

x + 1800 = 3100 + 1800

= $4900

$4900 is invested at 11%

Hope this helped.

6 0

3 years ago

At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine

Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

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 $s = \sqrt{\frac{p(1-p)}{n}}$