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coldgirl [10]
3 years ago
11

Amy is doing a science experiment on how a certain bacterium reacts to an antibiotic. She has 3 dishes of identical bacterium sa

mples with 14 bacteria in each dish. She gives an antibiotic to all of the bacteria in one dish. All of the treated bacteria died, and the bacteria in the other two dishes survived.
Is there a sampling bias in the situation above?
A.
No. All 3 dishes are filled with the same number of identical bacteria.
B.
Yes. The antibiotic may not work on the other bacteria.
C.
There is not enough information.
Mathematics
1 answer:
Kay [80]3 years ago
4 0

Answer:

the answer is A please give me hearts and stars because it is correct.

Step-by-step explanation:

it is easy you just mark out the ones you know that is wrong and you are left with the answer a which is very correct.

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Please help me its about Solving by substitution
Len [333]

Step-by-step explanation:

x=y+7

3x+2y=6

so wherever you see X you but y+7

3(y+7)+2y=6

3y+21+2y=6

group like terms

3y+2y=6-21

5y=-15

divide by 5

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4 0
3 years ago
Need help asap!!<br> First right answer will get brainliest
aleksklad [387]

answer:

B and C are correct.

explanation:

\hookrightarrow \sf x^2-5x+5  where ax² + bx + c

\\ Thus we can determine a = 1, b = -5, c = 5 \\

using quadratic equation:

\rightarrow \sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \  when \  \ ax^2 + bx + c = 0

\hookrightarrow \sf x_{1,\:2}=\dfrac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}

\hookrightarrow \sf  x=\dfrac{5+\sqrt{5}}{2},\:x=\dfrac{5-\sqrt{5}}{2}

4 0
3 years ago
Read 2 more answers
Geometry help needed
velikii [3]
  • Acute: less than 90 degrees
  • Right: equal to 90 degrees
  • Obtuse: greater than 90 degrees
  • Straight: equal to 180 degrees

4. m∠LGH = 14 ---acute

5. m∠SRT = 114 ---obtuse

6. m∠SLI = 90 ---right

7. m∠1 = 139 ---obtuse

8. m∠L = 179 ---obtuse

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7 0
4 years ago
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Select the correct answer from each drop-down menu.
Oksi-84 [34.3K]

Answer:

we \: have \: ohms \: law \: h =  {i}^{2} rt \\  \frac{h}{t}  =  {i}^{2} rt \div t = p \\  p =  {i}^{2} r....(1) \\ we \: have \: v = ir \\ i =  \frac{v}{r} .....(2) \\ put \: (2) \: in \: (1) \\ then \: p =  { (\frac{v}{r}) }^{2} r \\ p =  \frac{ {v}^{2} }{r}  \\ v =  \sqrt{pr}  =  {(pr)}^{ \frac{1}{2} }  \\ 2) \\ here \: r = 32 \: and \: p = .5 \\ then \: v =  \sqrt{32 \times .5}  \\  =  \sqrt{16}  = 4volt \\ thank \: you

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