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2 years ago

A triceratops has three horns How many terms do u think a trinomial has

Mathematics

1 answer:

larisa86 [58]2 years ago

4 0

The word 'tri' in 'trinomial' means three.

So, there are 3 terms in a trinomial.

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8 + 3 (16 - 12). Help thanks

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

you've got to do the inside parenthesis first then multiple with 3 and finally add 8

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2 years ago

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A 8-sided number die is rolled exactly once. Which statement is correct?

UkoKoshka [18]

Answer:

what statement if it is statistics it is 1/8 chance to get any number between 1-8

Step-by-step explanation:

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2 years ago

2tanx - 3cotx + 1 = 0

frutty [35]

Answer:

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Step-by-step explanation:

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2 years ago

Write 2 equivalent ratios for each ratio given:<br> 2/7<br> 3:3 1/2<br><br> 2.5 to 6

goldenfox [79]

Answer:

2/7 = 4/14

3:3 = 6:6

1/2 = 2/4

2.5:6 = 5:12

Step-by-step explanation:

For the first one, we can multiply the numerator and denominator of the fraction by 2 to get an equivalent fraction which is 4/14

For the second one we can multiply each side by 2 to get the ratio of 6:6

For the third one we can multiply the numerator and denominator by 2 to get 2/4

For the last one we can multiply each side by 2 to get 5:12

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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