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3 years ago

9

Idk how to do this 4x+y=1

1 answer:

ivolga24 [154]3 years ago

6 0

4x+y=1
4x+y-1=1-1
4x+y-1=0

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The stemplot below displays the arm spans, in centimeters, for 20 students.

Sindrei [870]

Answer:

189 and 202

Step-by-step explanation:

An outlier is a value that will have an effect on the mean of a set of values. It has an obvious difference compared to other values. They can be easily determined in a set of given data by mere observation.

From the given data in the question, it would be observed that the chronological difference among set of the data is either 1, 2, 3 or 4. But the two numbers that could be referred to as outliers are 189 and 202. This is because the have a wider range than others.

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2 years ago

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A tortoise is walking in the desert. It walks at a speed of

NARA [144]

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2 years ago

How many solutions does the system have?

Mrac [35]

Answer:

A Exactly one solution

Step-by-step explanation:

$\left \{ {{6x-3y=9} \atop {6x+y=9}} \right.\\\\\left \{ {{6x-3y=9} \atop {-6x-y=-9}} \right\\\\Then, \\\\-4y=0\\\\\\ y=0\\\\$

Putting y=0 at second equation, we have

$-6x=-9\\\\x=\frac{9}{6} =\frac{3}{2}$

The unique solution is $(\frac{3}{2} , 0)$

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2 years ago

A piece of string 25 and 1/2 inches long will be cut into 3/4 inch pieces how many pieces will there be

olga2289 [7]

you would have 9.375 pieces or 75/8 pieces, depends on how you look at it

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2 years ago

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

<em>P=0.0000037</em>

<em>P=0.00037%</em>

Step-by-step explanation:

<u>Probability</u>

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

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2 years ago