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katrin [286]
2 years ago
12

Consider these scenarios. 1. An elephant weighs 1. 5 × 104 units. 2. A mouse weighs 2. 13 × 10-5 units. 3. A puppy weighs 1. 2 ×

102 units. Determine the unit of measurement that best represents each scenario. 1. 2. 3.
Mathematics
1 answer:
Rudik [331]2 years ago
7 1

An ounce unit will best represent each scenario.

The average weight of an elephant is around 3000-4000 kg.

The average weight of a mouse = 15-20 grams

The average weight of a puppy = 25-30kg

It is given that

Weight of the elephant = 1.5 * 10^5 units ≈150000 units

Weight of the mouse = 6.3*10^2 units ≈630 units

Weight of the puppy = 1.2*10^3 units ≈1200 units

<h3>What is the relation between an ounce and a kg?</h3>

1 ounce = 0.0253495 kg

150000 ounces = 3800kg, near to actual value

630 ounces = 16kg, near to actual value

1200 ounces = 30kg, near to actual value

Therefore, an ounce unit will best represent each scenario.

to get more about units of a mass visit:

brainly.com/question/229459

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All you have to do is substitute x for 3.
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14.50 times 3 is 43.50

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Mr. Bennett owns 4 restaurants. The mean number of tables is 30. Which number line shows what could happen to the mean if he buy
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Number of tables al four restaurants have;
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This is the mean so it decreases. I hope this helps even though you didn’t actually post the number line
7 0
3 years ago
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Use the functions a(x) = 3x + 10 and b(x) = 2x − 8 to complete the function operations listed below.
Gnoma [55]
For part A, you simply add the two functions a(x) and b(x). The resulting function is then (a+b)(x) = 5x +2.

For part B, you multiply the functions a(x) and b(x). Using the FOIL method, we obtain, 6x^2 -24x + 20x - 80. Simplifying, we get, (a*b)(x) =  6x^2 -4x - 80.

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7 0
3 years ago
A process that fills packages is stopped whenever a package is detected whose weight falls outside the specification. Assume tha
swat32

Answer:

The mean number of packages that will be filled before the process is stopped is 50.

Step-by-step explanation:

For each package, there are only two possible outcomes. Either it fails outside the specifications, or it does not. Packages are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Number of trials expected for n sucesses

Also called inverse binomial distribution, is given by:

E = \frac{n}{p}

In which p is the probability of a success in a trial.

Assume that each package has probability 0.02 of falling outside the specification and that the weights of the packages are independent.

This means that p = 0.02

Find the mean number of packages that will be filled before the process is stopped.

This is when n = 1, as the process is stopped when a package is outside the specifications. So

E = \frac{n}{p} = \frac{1}{0.02} = 50

The mean number of packages that will be filled before the process is stopped is 50.

5 0
2 years ago
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Answer: C) 4(x+33)

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2. When you simplify it and add the like terms, you obtain:

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3. Now, you can factor out 4. Therefore, you can write the expression as below:

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4. So, you can conclude that the answer is the option C.

6 0
3 years ago
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