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trapecia [35]
3 years ago
13

3x < −2x + 7

Mathematics
1 answer:
Hunter-Best [27]3 years ago
8 0
Answer: B

Because...
3x + 2x < -2x + 2x + 7
5x < 7
So you have to do addition BEFORE division.
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josie enjoys making strawberry milkshakes. if it takes 5 strawberries yo make one milkshake how many does she need to make 20
Vinil7 [7]

Answer:

100 strawberries

Step-by-step explanation:

multipy 5 by 20 and you'll get 100

3 0
3 years ago
Read 2 more answers
13 please I have no idea what the steps for this is
Ilya [14]

Answer:

UV=142

Step-by-step explanation:

21x-13=10x+31

+13 +13

21x=10x+44

-10x=-10x

11x=44

÷11 ÷11

x=4

now substitute 4 in place of x (only have to solve one)

21(4)-13= 71

now multiply your answer by 2

71×2=142

6 0
3 years ago
Veronica and Chad are walking down the street. Veronica walks at a speed of 5 miles per hour, and Chad walks at a speed of 4 mil
stich3 [128]

Answer:

Veronica:

d=5t

Chad:

d=4t

Step-by-step explanation

yes they do represent linear equations

4 0
3 years ago
What is required in aerobic respiration?<br>​
likoan [24]
O2, or Oxygen, and C6H12O6, or glucose.
6 0
2 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
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