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3 years ago

Solve the system of equation by graphing

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

It is A because I saw your other post

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Help me for 1 through 6 pls I WANT TO FINISH ON TIME THANKS YOU

umka2103 [35]

1) put a bar over the last 4
2) put a bar over the last 3 numbers (198)
3) Put a bar over the last 2 numbers (39)
4)0.818181
5)0.199999
6)0.36743674

hope this helps

4 0

3 years ago

What is the value of the expression 2 superscript 5? 10 25 32 64

Varvara68 [4.7K]

32, that’s just 2 to the fifth power so 2*2*2*2*2

3 0

3 years ago

Read 2 more answers

Sarah went to the candy store to buy treats for her halloween party. She came out with 11 pounds of candy. She bought m&m's

Papessa [141]

Answer:

5lb of m&mns and 6lb of raisinets

Step-by-step explanation:

5lb x 2$ = $10

6lb x 1.5 = $9

8 0

3 years ago

If f(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align=

boyakko [2]

Answer:

Solution given:

f(x) = $\frac{1}{9}$ x-2

let f(x)=y

y = $\frac{1}{9}$ x-2

interchanging role of x and y

x= $\frac{1}{9}$ y-2

x+2= $\frac{1}{9}$ y

y=9(x+2)

y=9x+18

F-¹(x)=9x+18

and

F-¹(-1)=9*-1+18=-9+18=9

6 0

3 years ago

The amount of time, in minutes, that a woman must wait for a cab is uniformly distributed between zero and 12 minutes, inclusive

murzikaleks [220]

Answer:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

Step-by-step explanation:

Let X the random variable of interest that a woman must wait for a cab"the amount of time in minutes " and we know that the distribution for this random variable is given by:

$X \sim Unif (a=0, b =12)$

And we want to find the following probability:

$P(X$

And for this case we can use the cumulative distribution function given by:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

7 0

3 years ago