I’m stuck on g. and h.

Pani-rosa [81]

Vertical asymptotes can be found by setting the denominator equal to 0

x^2-6x = 0
factor out x
x(x-6)
set the two parts equal to 0
x = 0

x-6 = 0
x = 6

so the two x values are 0 and 6

ANSWER: B. x = 0, x = 6

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

1. The cafeteria manager at a middle school wanted to keep track of how many student breakfast and lunches were sold on a Monday

lianna [129]

I know for a fact that 1, 2 and 4 are correct. I'm almost positive 3 and 5 are right too

1. B

2.D

3.D

4.B

5.D

6 0

4 years ago

Is 0.1km greater than 10mm?

mixer [17]

$0.1km=0.1\cdot1000m=100m=100\cdot1000mm=100\ 000mm > 10mm$

5 0

3 years ago

Read 2 more answers

-403 - -177 ( show your work )

nika2105 [10]

<span>-403 - -177
Since two negative signs equal one positive sign, convert - -177 to +177
-403+177
I would make -403 into 403 and subtract 177 from it. Then at the end, put the negative sign next to the result.
226
Put the negative sign next to 226 since it was removed earlier from 403.
Final Answer: -226</span>

7 0

3 years ago

Read 2 more answers