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Sonbull [250]
3 years ago
12

Please help me solve each question! Make sure to show work so I can earn full credit.

Mathematics
2 answers:
omeli [17]3 years ago
8 0

(1)

5x-5\geq10\,\,\mbox{or}\,\,-3x+1>13\\5x\geq15\,\,\mbox{or}\,\,-3x>12\\x\geq3\,\,\mbox{or}\,\,x

(2)

5x + 3\leq18 \,\,\mbox{and}\,\,4 - x < 6\\5x\leq15 \,\,\mbox{and}\,\,- x < 2\\x\leq3 \,\,\mbox{and}\,\,x > 2\\2

IgorC [24]3 years ago
5 0

Answer:

1.  x < -4 or x ≥ 3

2.  -2 <  x ≤  3<em> </em>

Step-by-step explanation:

1. 5x - 5 ≥ 10

5x ≥ 15

x ≥ 3

-3x + 1 > 13

-3x > 12

x < -4

Solution is       x < -4 or x ≥ 3


2. 5x + 3 ≤ 18

5x ≤ 15

x ≤ 3

4 - x < 6

-x < 2

x > - 2

answer is   -2 < x ≤  3

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X less than or equal to 2
8 0
3 years ago
Which is greater 2/3 or 60%
zlopas [31]
2/3 because 2/3 is pretty much 66% 
6 0
3 years ago
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The blades of a windmill turn on an axis that is 35 feet above the ground. The blades are 10 feet long and complete two rotation
Alona [7]

Answer:

h = 10sin(π15t)+35

Step-by-step explanation:

The height of the blade as a function f time can be written in the following way:

h = Asin(xt) + B, where:

B represets the initial height of the blade above the ground.

A represents the amplitud of length of the blade.

x represents the period.

The initial height is 35 ft, therefore, B = 35ft.

The amplotud of lenth of the blade is 10ft, therefore A = 10.

The period is two rotations every minute, therefore the period should be 60/4 = 15. Then x = 15π

Finally the equation that can be used to model h is:

h = 10sin(π15t)+35

5 0
3 years ago
Read 2 more answers
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts
DochEvi [55]

Answer:

We accept  H₀   with the information we have, we can say level of ozone is under the major limit

Step-by-step explanation:

Normal Distribution

population mean  =   μ₀   =  7.5   ppm

Sample size     n   =  16      df  =  n  -  1  df  =  15

Sample mean      =    μ   =  7.8  ppm

Sample standard deviation   =  s  =  0.8

We want to find out if ozono level, is above normal level that is bigger than 7.5

1.- Hypothesis Test

null hypothesis                          H₀         μ₀   =  7.5

alternative hypothesis             Hₐ          μ₀  >  7.5  

2.-Significance level    α  =  0.01   we will develop one tail-test (right)

then   for   df  =  15    and    α =  0,01   from t -student table we get

t(c)  = 2.624

3.-Compute  t(s)

t(s)  =  (  μ -  μ₀ ) / s /√n            ⇒   t(s)  = ( 7.8  -  7.5 )*4/0.8

t(s)  = 0.3*4/0.8

t(s)  = 1.5

4.-Compare   t(s)   and  t(c)

t(s)  <  t(c)       1.5  <  2.64

Then  t(s) is inside the acceptance region. We accept  H₀

7 0
3 years ago
Someone help pls and thanks
Amiraneli [1.4K]
C I am soooo sorry if I’m wrong
4 0
3 years ago
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