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natali 33 [55]
2 years ago
8

If $x$ is the average of $13$, $-16$, and $6$ and if $y$ is the cube root of $8$, find $x^2 + y^3$. Please Help

Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
5 0

Answer:

If we have a set of N elements:

{x₁, x₂, ..., xₙ}

The mean value (also called the average value) Is calculated as:

Average = (x₁ + x₂ + ... + xₙ)/n

So if x is the average of 13, -16 and 6 ( a total of 3 values)

x will be equal to:

x = (13 + (-16) + 6)/3 = (19 - 16)/3 = 3/3 = 1

x = 1

And we know that:

y = ∛8

Remember that:

2*2 = 4

and

4*2 = 8

then

2*2*2 = 2^3 = 8

then ∛8 = 2.

So we have:

y = ∛8 = 2

Now we can replace these values in the equation:

x^2 + y^3

replacing:

x = 1

y = 2

we get:

1^2 + 2^3 = 1 + 8 = 9

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To produce at a point lying ______ the production possibilities curve would require economic growth.
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To produce at a point lying inside the production possibilities curve would require economic growth.

<h3>What is production possibilities curve ?</h3>

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2 years ago
George wishes to increase the percent of acid in 50 ml of a 15% acid solution in water
PtichkaEL [24]

Answer:

screw this question

Step-by-step explanation:

To change from 15% to 25% he needs to add 10%.

If the solution is 50ml then he needs to add 10% which is 5ml... But if he adds 5 ml then the solution will be 55ml total, so that doesn't make sense. He has to take some of the solution that's not acid before adding the 5ml...

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2 years ago
Read 2 more answers
$105 is what percent of $70?
REY [17]

Answer:

70$ =   150%   = 105 Divided by 1.5

Your answer is 150%

Another way of solving

105/70        0/100

Multiply 105 and 100= 10,500

then divide by 70 = 150

5 0
3 years ago
At 7:00 a.m., Alicia pours a cup of tea whose temperature is 200°F. The tea starts
soldier1979 [14.2K]

Answer:

Step-by-step explanation:

Here we have the temperature variation with time given in an exponential equation as follows;

Let the temperature at time x (in minutes) = y

Therefore, y = a × mˣ + b

Where:

b = Shift of the curve or the limit value of the decreasing exponential function as x → ∞

When y = 200, x = 0

Therefore. 200 = a × m⁰ + c = a + c

We note that c is the shift of the graph, the value upon which temperature increases = final temperature = 72°F

Hence a = 200 - 72 = 128°F

When y = 197, x = 2 minutes

Therefore, 197 = 128·m² + 72 =

m² = (197 - 72)/128 = 125/128

m = √(125/128) = 0.98821

Hence the exponential equation of cooling is presented in the following equation;

y = 128 × (0.98821)ˣ + 72

Therefore, y = 128(0.989)x + 72

When the temperature is 172°F we have;

170 = 128 × (0.989)ˣ + 72

∴ (0.989)ˣ = (170 - 72)/128

= 98/128

= 0.7656

log(0.989)ˣ = log(0.7656)

x·log(0.989) = log(0.7656)

x = log(0.7656)/log(0.989)

= 24.15 minutes

5 0
3 years ago
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