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3 years ago

Cuanto me da como resultado

Mathematics

1 answer:

Likurg_2 [28]3 years ago

4 0

Step-by-step explanation:

1) $(-4)^2{\cdot}(-4)$

We know that, $a^x{\cdot}a^y=a^{x+y}$

$(-4)^2{\cdot}(-4)=(-4)^{2+1}\\\\=(-4)^3\\\\=64$

2) $(-2)^5{\cdot} (-2)^3$

Again using above property,

$(-2)^5{\cdot} (-2)^3=(-2)^8\\\\=256$

3) $5^{-3}$

We know that,

$a^{-x}=\dfrac{1}{a^x}$

So,

$5^{-3}=\dfrac{1}{5^3}\\\\=\dfrac{1}{125}$

4) $2.5^2=2.5\times 2.5\\\\=6.25$

Hence, this is the required solution.

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There are 3 times more kittens than puppies in a box. Together there are 24 animals. How many kittens are there?

Paha777 [63]

kitten : puppies

3:1

Total no.

3+1 = 4

Therefore

24÷4 = 6

(each of ratio represent 6 animals)

So,

3×6

= 18

Answers

18 kitten

6 0

3 years ago

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft and on the other three s

makvit [3.9K]

Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = $Length \times Breadth = xy$

We are given that the area of the garden is 122 square feet

So, $xy=122$ ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

Substitute the value of y from A

Total cost = $30x+20(\frac{122}{x})$

Total cost = $\frac{2440}{x}+30x$

Now take the derivative to minimize the cost

$f(x)=\frac{2440}{x}+30x$

$f'(x)=-\frac{2440}{x^2}+30$

Equate it equal to 0

$0=-\frac{2440}{x^2}+30$

$\frac{2440}{x^2}=30$

$\sqrt{\frac{2440}{30}}=x$

$9.018 =x$

Now check whether it is minimum or not

take second derivative

$f'(x)=-\frac{2440}{x^2}+30$

$f''(x)=-(-2)\frac{2440}{x^3}$

Substitute the value of x

$f''(x)=-(-2)\frac{2440}{(9.018)^3}$

$f''(x)=6.6540$

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

$(9.018)y=122$

$y=\frac{122}{9.018}$

$y=13.528$

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

4 0

3 years ago

Fred the ant is on the real number line, and Fred is trying to get to the point $0.$

Pavlova-9 [17]

Complete question is;

Fred the ant is on the real number line, and Fred is trying to get to the point 0

If Fred is at 1 then on the next step, Fred moves to either 0 or 2 with equal probability. If Fred is at 2 then on the next step, Fred always moves to 1

Let e1 be expected number of steps Fred takes to get to 0 given that Fred starts at the point 1. Similarly, let e2 be expected number of steps Fred takes to get to 0 given that Fred starts at the point 2

Determine the ordered pair (e1, e2)

Answer:

(e_1,e_2) = (2,3)

Step-by-step explanation:

We can track the probabilities using the scenario that Ant gets to 0 after 1 steps, 2 steps, 3 steps, 4 steps, 5 steps and so on.

This gives us e_1 = 1(1/2^(1)) + 2(1/2²) + 3(1/2³) + 4(1/2⁴) + ...

By arithmetico-geometric series, which is given by;

S_(∞) = [a/(1 - r)] + dr/(1 - r)²

From the online calculator, i got;

e_1 = 2

Similarly, e_2 = 2(1/2^(1)) + 3(1/2²) + 4(1/2³) + 5(1/2⁴) + ...

By arithmetico-geometric series, e_2 = 3,

Thus, (e_1,e_2) = (2,3).

6 0

3 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0

4 years ago

I need help and it’s due today

olga nikolaevna [1]

Answer: same bruh

Step-by-step explanation:

7 0

4 years ago