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fredd [130]
3 years ago
15

Help ASAP I’m bad at math lol,Solve for d

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
8 0

it is -16 because I took the test

Leya [2.2K]3 years ago
7 0
The answer is -16!!!
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HHEELLPP MMEE!!!!!!!!!!!
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Since ALL the data from the weekdays are lower than the weekends, that means the fewer cars were sold during the week than on weekends.
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Imagine that you take a road trip from A to D, but you have to do it in segments. Let’s say the distance from A to B is 145 mile
Rzqust [24]

Answer: The average is 70.

Step-by-step explanation:

145+160+115=420

420 divided by 6 =70

So the average is 70.

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What is the range of the function in this table ?
steposvetlana [31]

Answer:

B

Step-by-step explanation:

Range refers to all the y values that a function has. The table shows that the function has y values: 1, 2, and 4.

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4 years ago
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Let f be an exponential function and g be a logarithmic function.
love history [14]

Good evening ,

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Answer:

1) (f+g)(1) = e

2) (fg)(1) = 0

3) (3f)(1) = 3e

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Step-by-step explanation:

1) (f+g)(1) = f(1) + g(1) = e¹ + log1 = e + 0 = e.

2) (fg)(1) = f(1) × g(1) = e¹ × log(1) = e¹ × 0 = e × 0 = 0.

3) (3f)(1) = 3×f(1) = 3×e¹ = 3e.

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8 0
4 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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