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3 years ago

Is (4•6)^3 equivalent to 4•6^3

Mathematics

2 answers:

Vinil7 [7]3 years ago

5 0

Answer:

No, (4•6)^3 is not equivalent to 4•6^3

Step-by-step explanation:

(4•6)^3

4•6^3

Let's solve each one at a time, starting with (4•6)^3

(4•6)^3

Multiply inside the parenthesis first

24^3

Multiply 24 by itself 3 times (Exponents)

24 * 24 * 24

13,824

Now let's solve 4•6^3

4•6^3

Exponents come before multiplication in the order of operations, so multiply 6 by itself 3 times.

4 * (6x6x6)

4 * 216

864

Now we can compare the solutions

13,824 > 864

No, 13,824 is not equivalent to 864.

Hope this helps :)

Pani-rosa [81]3 years ago

4 0

No because the exponent on 6^3 is 216 and if you use (4•6)^3 then you’ll have to multiply the quotations and then Square it to the third

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r-ruslan [8.4K]

Answer:

Domain: (−∞,∞),{x|x∈R}

Range: (−∞,∞),{y|y∈R}

Step-by-step explanation:

Find the domain by finding where the function is defined. The range is the set of values that correspond with the domain.

Domain: (−∞,∞),{x|x∈R}

Range: (−∞,∞),{y|y∈R}

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2 years ago

what is the point-slope form of the equation of the line passing through the points (–6, –4) and (2, –5).

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Answer:

The equation of the line would be y + 4 = -1/8(x + 6)

Step-by-step explanation:

To find the point-slope form of the line, start by finding the slope. You can do this using the slope formula below along with the two points.

m(slope) = (y2 - y1)/(x2 - x1)

m = (-4 - -5)/(-6 - 2)

m = 1/-8

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Now that we have the slope, we can use that along with one of the points in the base form of point-slope.

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3 years ago

Solve for x.

liubo4ka [24]

Answer:

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2 years ago

Read 2 more answers

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

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S_A_V [24]

Answer:

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Step-by-step explanation:

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