[[|help please|]] will give brainliest

Could a square with whole number side lengths have the same perimeter as the dog pen? The same area

Jobisdone [24]

Answer:yes

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What integer is equivalent to 9^3/2

ella [17]

Here is the problem and amswer 9/3=3 3/2= 1.5

4 0

3 years ago

suppose you know that over the last 10 years, the porbabiliy that your town would have at least one ma jor stor was 40%. describ

Anuta_ua [19.1K]

Answer:

The probability of a big storm is 40%.

Now, you can find a D10 (a die with 10 faces)

You can assign 4 numbers (0, 1, 2, 3) to the event "there is a major storm"

in this way, you have 40% of having a storm.

and the other 6 numbers (4, 5, 6, 7, 8, 9) to the event "there is not a major storm". This means that we have 60% of not having a storm.

now, when you roll the dice you can see if a year there will be a storm or not.

Now, you roll the dice 5 times (for the 5 years) and take note of the results and the number of storms in those 5 rolls.

Now do the same thing a bunch of times, at least 25 times.

Now, you recorded the results in each set of rolls, now see the number of sets that have at least years with storms.

Take that number and divide it by the total number of sets of data (in this case 25, for example)

The result is the probability that we are looking for.

6 0

3 years ago

A person making $500 a week receives a 12% raise, which is an increase in their salary. Find the person’s new weekly salary.

Murrr4er [49]

500 times 1.12 or 12% equals 560

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

7 0

3 years ago