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lorasvet [3.4K]
3 years ago
11

URGENT PLEASE!!!!!!!!!!!!

Mathematics
2 answers:
blondinia [14]3 years ago
8 0

Answer:

its x=16 :)))

kondaur [170]3 years ago
6 0

Answer:

i got 178

Step-by-step explanation:

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Which function is graphed?
Alexandra [31]

Hey there! :)

Answer:

C. {x² + 4, x < 2

    {-x + 4   x ≥ 2

Step-by-step explanation:

This is a piecewise function, where the two equations are different. They are:

y = x²+ 4

y = -x + 4

The function x² + 4 is graphed where x < 2. (< is used because the circle is open)

The function -x + 4 is graphed where x ≥ 2. (≥ is used since the endpoint is closed)

Therefore, the correct answer is:

C. x² + 4, x < 2

   -x + 4   x ≥ 2

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3 years ago
Which factors can be multiplied together to make the trinomial 5x2 + 8x – 4? Check all that apply. (x + 1) (2x + 1) (x + 2) (5x
andreyandreev [35.5K]

Answer:

(5x-2) and (x+2)

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
Fraction 13/250 as a decimal
Luba_88 [7]
To find the decimal form of 13/250 you must divide 13 by 250 which is 0.052
5 0
4 years ago
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Brandon had $500. He spent 47% of his money on a tablet computer. How much did his computer cost?
Kobotan [32]
47% of 500 is 235 dollars
8 0
3 years ago
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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
AURORKA [14]

Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)

Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

7 0
3 years ago
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