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3 years ago

Solve the graphs? Show Your Work?

Mathematics

1 answer:

kari74 [83]3 years ago

4 0

$4.\\x\leq10\\\\5.\\x\leq-4\\\\6.\\x\geq4$

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Denzel is comparing web sites for downloading music.one charge a$5membership fee plus $0.5 per song .another charges $1.00 per s

MA_775_DIABLO [31]

.5x+$5=$1x
x=11
.5*11+5=11
1*11=11

4 0

4 years ago

Put the equation in standard form with a positive X coefficient. y-5 = -2 (x+6)

iVinArrow [24]

The answer is 2x+y+7=0.

3 0

4 years ago

Someone help me please I don’t get it

rodikova [14]

Answer:

x = $\frac{15}{2}$ , x = 1

Step-by-step explanation:

(a)

$\frac{4x-1}{2}$ = x + 7 ( multiply both sides by 2 to clear the fraction )

4x - 1 = 2x + 14 ( subtract 2x from both sides )

2x - 1 = 14 ( add 1 to both sides )

2x = 15 ( divide both sides by 2 )

x = $\frac{15}{2}$

(b)

3x + 2 = $\frac{2x+13}{3}$ ( multiply both sides by 3 to clear the fraction )

9x + 6 = 2x + 13 ( subtract 2x from both sides )

7x + 6 = 13 ( subtract 6 from both sides )

7x = 7 ( divide both sides by 7 )

x = 1

8 0

3 years ago

The shuttle bus from your parking lot and your office building operates on a 15 minute schedule. You arrive at the parking lot a

mylen [45]

Answer:

(a) The standard deviation of your waiting time is 4.33 minutes.

(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.

Step-by-step explanation:

Let X = the waiting time for the bus at the parking lot.

The random variable X is uniformly distributed with parameters a = 0 to b = 15.

The probability density function of X is given as follows:

$f_{X}(x)=\frac{1}{b-a};\ a$

(a)

The standard deviation of a Uniformly distributed random variable is given by:

$SD=\sqrt{\frac{(b-a)^{2}}{12}}$

Compute the standard deviation of the random variable X as follows:

$SD=\sqrt{\frac{(b-a)^{2}}{12}}$

$=\sqrt{\frac{(15-0)^{2}}{12}}$

$=\sqrt{\frac{225}{12}}$

$=\sqrt{18.75}\\=4.33$

Thus, the standard deviation of your waiting time is 4.33 minutes.

(b)

The value representing 2 standard deviations is:

$X=2\times SD=2\times4.33=8.66$

Compute the value of P (X > 8.66) as follows:

$P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx\\$

$=\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx\\$

$=\frac{1}{15}\times |x|^{15}_{8.66}\\$

$=\frac{15-8.66}{15}\\$

$=0.4227$

Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.

4 0

3 years ago

Write the numbers from the least to the greatest

Anvisha [2.4K]

I'm pretty sure the answer is b. because obviously 1.2 in is the least and the second least is the square root of 3.

8 0

3 years ago