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Mrac [35]
2 years ago
7

10ⁿ = 1 a) n=1 b) n=0 c) n=2

Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
6 0

Step-by-step explanation:

n = 0

because,

p^0 = 1

10000000^0 = 1

Georgia [21]2 years ago
4 0
B) n = 0

Anything to the power of 0 not matter the number is always equal to 1.

For ex:

1^0 = 1
100^0 = 1
10000000000^0 = 1

Even 0^0 = 1

Hope that helps (:
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Which factorization could represent the number of water bottles and weight of each water bottle?
motikmotik

The factorization that could represent the number of water bottles and weight of each water bottle is 12(5x^2 + 4x + 2). Option B

<h3>What is factorization?</h3>

The term factorization has to do with the process of obtaining common factors in an expression. It involves dividing each term in the expression with a factor that is common to all the terms in the expression.

The factorization that could represent the number of water bottles and weight of each water bottle is 12(5x^2 + 4x + 2).

Learn more about factorization:brainly.com/question/19386208

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Missing parts;

Mara carried water bottles to the field to share with her team at halftime. The water bottles weighed a total of 60x2 + 48x + 24 ounces. Which factorization could represent the number of water bottles and weight of each water bottle? 6(10x2 + 8x + 2) 12(5x2 + 4x + 2) 6x(10x2 + 8x + 2) 12x(5x2 + 4x + 2)

8 0
2 years ago
The profit function of a manufacturer, in dollars, when x units of a certain commodity are produced and sold is given by
Annette [7]

Answer:

Derivative: P’(x) = 200 - 2x

If P’(x) = 0 , x = 100

Therefore P’ is positive and P increases until x = 100

So the maximum units to maximize profits is 100 units

And for 100 units the profit in dollar is:

20000 - 10000 = 10000

10000 dollars

4 0
3 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0
2 years ago
Solve the following inequality.<br><br> 9x−2&gt;43
kirill [66]

Answer:

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3 years ago
5.7 as a mixed number in simplist form
Sveta_85 [38]
5.7 is 5 7/10 and it cannot be simplified
6 0
2 years ago
Read 2 more answers
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