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3 years ago

Please help me !

Mathematics

2 answers:

Alik [6]3 years ago

8 0

Solve for x in the second equation first

kumpel [21]3 years ago

8 0

Answer:

Solve for x in the second equation.

Step-by-step explanation:

To solve a system of equations by substition you have to solve one of the equations for X or Y and then substitute that value on the X of the other equation, if you want to avoid fractions, you can use the x of the secon equation, like this:

$6x+18y=24\\6x=24-18y\\x=\frac{24-18y}{6} \\x=4-3y$

So by solving X for the second equation you avoided fractions when substituting the value for X on the first equation.

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64 is 4 times the difference between Sarah's age, a, and 44. Assume Sarah is older than 44.

nlexa [21]

Answer:

Sarah is 60 years old

Step-by-step explanation:

64 divided by 4 is 16. Add 16 to 44 to get Sarah's age, 60

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2 years ago

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How do you write 30.0 in scientific notation? ____× 10^_____

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Answer:

It's written as

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3 years ago

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2 years ago

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Drew has 50 red, green, and blue marbles. He has six more red marbles than blue marbles and four fewer green marbles than blue m

Yanka [14]

Drew has 56 Red Marbles, 56 Green Marbles, Drew has about 60 Blue Marbles. First i added 6 to Red since Red Has 6 More Marbles Than Blue, Then i added Subtracted 4 from green, which resulted in 46 Green Marbles. Finally i added 4 and 6 to get 60 Blue Marbles. I am not sure if i am correct, but i gave it a shot! ^-^

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2 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