Please solve kx-2=7 for x

Please help. Ella has 5 pink bows, 1 blue bow, and 4 purple bows in a box, she will randomly choose 1 bow from the box. What is

pogonyaev

Answer:

4/10 chance. Or 2/5. 40% chance.

Step-by-step explanation:

6 0

3 years ago

5/ 2 x ≤ -14 A. -10.3 B. -6 C. -5.6 D. -5

Advocard [28]

Answer:

C

Step-by-step explanation:

5/2x<_-14

5x<_-28

5x/5<_-28/6

X<_-5.6

4 0

2 years ago

Read 2 more answers

Using the order of operations, what would be the first step: 4−8÷1−3[3+(7−5)2]

jeka94

Answer:

(7-5)

Step-by-step explanation:

According to PEMDAS, parenthesis go first. But there are two sets of parenthesis so the one inside the other goes first.

Next would be [3+(2*2)]

Then, [3+4]

Then, -8/1

Then, -3(7)

So the question would be 4-8-21

Which gives you -25 as an answer.

I realize that you weren't asking for the whole thing but maybe there are follow up questions.

3 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,

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