Parallelogram ABCD is a rectangle. What are the slopes of the sides that make this quadrilateral a rectangle?

Since nobody answered my other questions, can someone please show me how to solve and answer 14 and 15 all the parts? Please I n

ozzi

Sure : Do 3,000 x 6 and you will see what you get and that’s the answer

3 0

3 years ago

In college basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points ins

denis-greek [22]

Answer:

a) By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, 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 proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.45, n = 100$

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

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

By the Central Limit Theorem

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

$Z = \frac{0.25 – 0.45}{0.0497}$

$Z = -4.02$

4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered

surprising, so yes, it would be surprising for him to make only 25 of these shots.

3 0

3 years ago

What's the value of the expression below:

elena-14-01-66 [18.8K]

Answer:

The answer is 5 5/8.

The correct answer is not in the options

3 0

3 years ago

Read 2 more answers

Look at the number line below.

VikaD [51]

Answer: Choice C)

We start at 0 and move 1.4 units to the right. Then we move another 2.3 units to the right to land on 3.7

Notice how 1.4 - (-2.3) = 1.4 + 2.3 = 3.7

5 0

3 years ago

Read 2 more answers

The batteries from a certain manufacturer have a mean lifetime of 850 hours, with a standard deviation of 70 hours, assuming tha

matrenka [14]

To answer (a), we will need to find the Z value for 710 hours and 990 hours.

$\begin{gathered} For\text{ 710} \\ Z\text{ = }\frac{x-\mu}{\sigma} \\ \text{ = }\frac{710-850}{70} \\ \text{ = -2} \\ p\text{ =0.0228 } \\ For\text{ 990:} \\ Z\text{ = }\frac{990-850}{70} \\ \text{ =2} \\ p\text{ = 0.9772} \end{gathered}$ $\begin{gathered} To\text{ find P \lparen–2 \le Z \le 2\rparen} \\ =\text{ 0.9772 -0.0228} \\ =\text{ 0.9544} \\ =95.44\% \end{gathered}$

B) 68% of data lies between one standard deviation of the mean.

850 +70 = 920

850 - 70 = 780

5 0

1 year ago