Identify the dependent variable in the situation. Coat sales go up when it is cold outside.

If a cup of ice cream weighs 1/4 pound how many dishes can u get from 4 1/2 pound?

Karolina [17]

4 1/2 divided by 1/4 9/2 x 4/1 = 36/2 18/1 = 18 18 cups of ice cream

5 0

3 years ago

Read 2 more answers

14x+10y=72 14x+4y=96 Solve the following system of equations using any method

Anarel [89]

  14x + 10y = 72
  14x +   4y = 96
  6y = -24
6 6
  y = -4
  14x + 10y = 72
14x + 10(-4) = 72
14x - 40 = 72
+ 40 + 40
  14x = 112
14 14
x = 8
  (x, y) = (8, -4)

5 0

3 years ago

Find the 100th term for the sequence. -9,-4, 1, 6, 11, .......

Paraphin [41]

Answer:

486

Step-by-step explanation:

for the 2nd term, add 5 to -9

for the 3rd, add 5 × 2 to -9

for the 4th, add 5 × 3 to -9

:

for the 100th, add 5 × (100 - 1) to -9

=> 5 × 99 + (-9)

=> 486

8 0

2 years ago

Given the arithmetic sequence an = 4 − 3(n − 1), what is the domain for n?

Oxana [17]

Answer:

The answer is All integers where n ≥ 1

Hope this helps!

8 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