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Licemer1 [7]
3 years ago
11

Solve the system of inequalities: 3x-10<0 2x>0 HELPPP WILL MARK BRAINLIEST

Mathematics
2 answers:
Pachacha [2.7K]3 years ago
7 0

Answer: <0<x<10/3

Step-by-step explanation:

son4ous [18]3 years ago
4 0

Answer:

<u>Solve each inequality and combine solutions:</u>

  • 3x - 10 < 0 ⇒ 3x < 10 ⇒ x < 10/3
  • 2x > 0 ⇒ x > 0

<u>Combination of the two is:</u>

  • 0 < x < 10/3

or

  • x ∈ (0, 10/3)
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Graph a line with a slope of 3/4 that contains the point (2,-3).
arsen [322]

Answer:

Step-by-step explanation:

First, plot the point (2, -3) on the graph.  Then, use the slope to pick another point.  The slope is rise over run.  For your slope, the line will go 3 places up and then 4 places to the right.  Using a straight-edge, follow the points and you will get the graph of the line.

6 0
3 years ago
Describe how the range of a data set can help describe its variability.
katovenus [111]

Answer:

The range is defined as the difference between the term with the highest value and the term with the lowest value. This statistic is used to measure the variability of a series of data because it provides information on how far apart the values of a tail of the distribution are from the values at the other end of the tail.

Imagine that you manufacture a type of spare part for cars that must have a measurement of 10 cm with a margin of error of 1 cm.

This is:

10 ± 1 cm

Then you expect your manufacturing process to produce pieces with identical dimensions, that is, with little variability.

If you randomly select a sample of n pieces and measure them, the variability is expected to be low, so that your process is of quality, then expect a low range preferably less than 1 cm.

{10, 10.1, 10.5, 9.8, 9,6, 10.2} Range= 10.5 - 9.6 = 0.9 cm <em> low variability</em>

But if you find that the range is up to 8 cm, this would mean that not all pieces measure around 10 cm, it means that the variability of the measurements is high.

{14, 12, 11, 8, 7, 11, 12, 15} Range = 15 - 7 = 8 cm   <em>high variability </em>

6 0
3 years ago
Read 2 more answers
Can anyone help me out, if you answer the question right i will give you a brainliest.
melisa1 [442]

Answer:

It's different because the experiment is more accurate as it progresses.

Step-by-step explanation:

You'll notice that the higher the numbers get in the experiment the closer it gets to your solution. The theoretical probability of flipping a coin is about 50% heads and 50% tails, but it doesn't always seem like that in an experiment. The experimental probability from your experimentation so far would be 62% of heads and 38% of tails.

4 0
3 years ago
Veterinarians often use nonsteroidal anti-inflammatory drugs (NSAIDs) to treat lameness in horses. A group of veterinary researc
ra1l [238]

Answer:

C. The chance to be selected into the sample was the same for all veterinarians

6 0
3 years ago
Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

C(t) = C(0)e^{rt}

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So C(0) = 110

Husband

Half life of 4 hours. So

C(4) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{4r}

e^{4r} = 0.5

Applying ln to both sides

\ln{e^{4r}} = \ln{0.5}

4r = \ln{0.5}

r = \frac{\ln{0.5}}{4}

r = -0.1733

So for the husband

C(t) = 110e^{-0.1733t}

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.1733t}

C(11) = 110e^{-0.1733*11} = 16.35

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

C(10) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{10}

e^{10r} = 0.5

Applying ln to both sides

\ln{e^{10r}} = \ln{0.5}

10r = \ln{0.5}

r = \frac{\ln{0.5}}{10}

r = -0.0693

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.0693t}

C(11) = 110e^{-0.0693*11} = 51.33

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0
4 years ago
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