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3 years ago

Perform the indicated operation and simplify show work please:

Mathematics

2 answers:

NARA [144]3 years ago

3 0

$\frac{x+4y}{x+y}+\frac{2x-y}{x+y}=\frac{x+4y+2x-y}{x+y}=\frac{3x+3y}{x+y}=\frac{3(x+y)}{x+y}=3\ ;(x \not= -y)$

stepan [7]3 years ago

3 0

x + 4y + 2x - y = x + 2x + 4y - y = 3x + 3y = 3
x + y x + y x + y x + y

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Whats 12y=9x-30 in standard form

nadya68 [22]

Standard form : Ax + By = C

12y = 9x - 30
-9x + 12y = -30 ....some teachers accept this, with A being negative
but if not, then it is this :
9x - 12y = 30

7 0

3 years ago

Can u help solve this

Klio2033 [76]

Answer:

3 or 6/2

Step-by-step explanation:

4--2 (you add there is a subtraction of a negative) over or divided by 3-1

4 0

3 years ago

Help how to solve! 24=3r divided by 4

zhenek [66]

24=3r/4

Multiply both sides by 4

24=(3r)/4

(24)*(4)=(3r/4)*(4)

96=3r

Flip the equation

3r=96

Divide both sides by 3

3r/3=96/3

r=32

Now lets check if the answer is good

24=3(32)/4

24= 96/4

24=24

You see , the answer is good .

I hope that's help ! Let me know .

6 0

3 years ago

.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and

Leni [432]

Answer:

a) Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

5 0

3 years ago

Simplify: (4/2d^2)^3

cestrela7 [59]

8/d^6

it needs to be 20 characters long so

7 0

3 years ago

Read 2 more answers