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3 years ago

If 3g - 11 + 3y, what is the value of y when g is 12?

Mathematics

2 answers:

Hitman42 [59]3 years ago

6 0

Answer:

y=25

Step-by-step explanation:

I think this may be the answer but I'm not sure.

Aneli [31]3 years ago

5 0

Answer:

25 + 3y

Step-by-step explanation:

3g - 11 + 3y when g = 12

~Substitute

3(12) - 11 + 3y

~Simplify

36 - 11 + 3y

~Combine like terms

25 + 3y

Best of Luck!

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Help me pleaseeeeeeeeeeeeeee

dangina [55]

Answer:

W=1000C/tc

Step-by-step explanation:

First multiply both side 1000: 1000C=Wtc

Divide both side by tc: 1000C/tc=W (t, c ≠0)

7 0

3 years ago

Help me pls

expeople1 [14]

I think you can find it graphically by points through a table you create like:

let x = 0, and find the value of y

for example for the first equation when we let x = 0 you find y = 6 and be in the form of (0,6) then draw them or you can let x = -2 or -3 , etc as you like

to be more accurate you can find 3 pionts and start graphing them

these two equations are straight lines and they will intersect in a point

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%7B75%7D%5E%7B2%7D%20%2B%2025%20-%20750" id="TexFormula1" title=" {75}^{2} + 25 - 750" alt=

frez [133]

your answer is 4900

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,

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

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

A distribution of values is normal with a mean of 1986.1 and a standard deviation of 27.2.

m_a_m_a [10]

Answer:

I used the function normCdf(lower bound, upper bound, mean, standard deviation) on the graphing calculator to solve this.

Lower bound = 1914.8
Upper bound = 999999
Mean = 1986.1
Standard deviation = 27.2

Input in these values and it will result in:

normCdf(1914.8,9999999,1986.1,27.2) = 0.995621

So the probability that the value is greater than 1914.8 is about 99.5621%

I'm not sure if this is correct 0_o

4 0

3 years ago