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3 years ago

Three to the sixth power divided by three to the third power

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

3 0

let's first convert this to number form.

3^6 / 3^3= what?

to divide a number with the same base but different exponent, we simply subtract the second exponent by the first exponent.

3^6 / 3^3 = 3^6-3=3^3

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Which transformation does not preserve orientation? A. rotation B. reflection across the y-axis C. dilation D. translation

Sophie [7]

The correct answer is B) reflection across the y-axis

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3 years ago

how much must you deposit in an account that pays 8% interest, compounded monthly, to have a balance of $1000 after 3 years?

Makovka662 [10]

Answer:

1000=N(1 + .08/12)^36 where N is the initial deposit.

So 1000=N(1.00666667)^36

N=1000/(1.00666667)^36

ln N=ln 1000/(1.00666667)^36=ln 1000-36 ln (1.00666667)

ln N=6.6685517411100682158933244766027

N=e^6.6685517411100682158933244766027=$787.25463 as the initial deposit.

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3 years ago

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.

6 0

3 years ago

How do I work this out? (Picture is included) q3

Julli [10]

3. 2+2=4, 2+4=6, 6+4=10, 10+6=16, 16+10=26, 26+16=42

the answer is 2, 2, 4, 6, 10, 16, 26, 42

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3 years ago

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spin [16.1K]

Answer:

Here you are. I hope this helps.

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3 years ago

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