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2 years ago

3n at n = 11 A) 8 B) 14 C) 7 D) 33

Mathematics

1 answer:

Aleks [24]2 years ago

3 0

<h3>Answer: D) 33</h3>

Explanation:

3n means "3 times n" where n is a placeholder for a number. Specifically, that number is 11. So,

3n = 3*n = 3*11 = 33

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H2 + 7h + 10 = 0<br> Help please

Leni [432]

Answer:

-2, -5

Step-by-step explanation:

h2 + 7h + 10 = 0

h^2 +7h + 10 = 0

h^2 +2h +5h + 10 = 0

h(h+2) + 5(h+2) = 0

h+2 = 0 or h+5 = 0

h = -2 or h = -5

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3 years ago

The table shows the average number of pounds of trash generated per person per day in the United States from 1970 to 2010 use th

stepan [7]

Answer: Mean=4.11

Median=4.44

Step-by-step explanation:

Using a graphing calculator, enter the values into the first list. Then run one-variable statistics on the list.

By hand, to find the mean, add together all of the values:

3.25+3.25+3.66+3.83+4.57+4.52+4.74+4.69+4.44 = 36.95

Divide by the total number of data points, 9:

36.95/9 = 4.11

To find the median, order the data from least to greatest:

3.25, 3.25, 3.66, 3.83, 4.44, 4.52, 4.57, 4.69, 4.74

The middle number is 4.44

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3 years ago

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Help me please i need help on my maths hw people

Nookie1986 [14]

65-14=65-5-x
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Explanation=
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2 years ago

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PLEASE ANSWER<br><br> What is a ratio?

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A ratio is the quantitative relation between two amounts showing the number of times one value contains or is contained within the other.

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2 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.

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3 years ago

3n at n = 11 A) 8 B) 14 C) 7 D) 33​

3n at n = 11 A) 8 B) 14 C) 7 D) 33