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Arlecino [84]
3 years ago
8

Graph y= 1/2x–3. (Ill mark brainliest)

Mathematics
2 answers:
11111nata11111 [884]3 years ago
8 0

Answer:

Slope is 1/2 and y intercept is -3 so it is a graph like this

Mekhanik [1.2K]3 years ago
3 0

Answer:

Step-by-step explanation:

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Compute the standard deviation of the following set of data to the nearest whole number: 8, 12, 15, 17, 18
attashe74 [19]
We have that

set of data [<span>8, 12, 15, 17, 18]

step 1
Find the mean
</span>[8+ 12 + 15 + 17 + 18]/5-------> 70/5------> 14

step 2
 for each number: subtract the Mean and square the result
(8-14)²----->36
(12-14)²----> 4
(15-14)²----> 1
(17-14)²----> 9
(18-14)²----> 16

step 3
<span> work out the mean of those squared differences
[36+4+1+9+16]/5-----> 66/5-----> 13.2----> </span><span>this value is called the "Variance"

step 4
</span><span>Take the square root of that
</span>standard desviation=√13.2-----> 3.63-------> 4

the answer is
4
7 0
3 years ago
Find x round to the nearest tenth. Hint: use the quadratic formula
Fofino [41]
What are the options
5 0
4 years ago
10 Mason's can build a wall working 7 hours a day,in 12 days.In how many days can the work be completed by 14 persons working 8
Ronch [10]

Answer:

7 hr * 12 days = 84 hours

10*84 = 840 man-hrs to build the wall

----

14 men * 8 hr = 112 man-hrs/day

840/112 = 7.5 days

Step-by-step explanation:

5 0
3 years ago
You have worked as a guide at the aquarium for two summers. This summer you have been offered the job again, but with a 4 percen
gulaghasi [49]

Answer:

ok i did 1875/2.5 months and got 750 then i took 750 and multiplied it by .04 and got 30

so 750+30 =780

780*2.5= 1950

so you earned 30 more dollars each month so the entire summer you worked you got $1950

Step-by-step explanation:

4 0
3 years ago
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
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