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SSSSS [86.1K]
3 years ago
11

A 9 cm tall cone shaped paper cup can hold up to 58.9 cm cubed of water. What is the minimum amount of paper needed to make the

paper cup, assuming no overlap in the paper? Use 3.14 for π.
Mathematics
1 answer:
babymother [125]3 years ago
4 0
Volume of a cone, V = πr^2h/3

r = Sqrt 3V/πh = Sqrt [(58.9*3)/(3.14*9)] = 2.5 cm, minimum

Surface area, A = πr [r+ Sqrt (h^2+r^2)] =  3.14*2.5*[2.5+ Sqrt (9^2+2.5^2)] = 92.95 cm^2

Therefore, minimum amount of paper required is 92.95 cm^2
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Find the area of a triangle with a base of 9 feet and hypotenuse length of 13 feet​
iris [78.8K]

Answer:

42.21 ft^2

Step-by-step explanation:

The formula: A=1 /2b √ c2﹣b2

c=hypotenuse

b=base

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3 years ago
I need help asap! due tomorrow and am clueless!!
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4 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
A sample of a radioactive isotope had an initial mass of 860 mg in the year 2010 and decays exponentially over time. A measureme
katen-ka-za [31]

Answer:

y=166

Step-by-step explanation:

7 0
3 years ago
Need help on question 4??
Anit [1.1K]

HK consists of two part, HJ and JK

HK = HJ + JK

We know that:

HK = 12.2ft

JK = 3.1ft

HJ = ?

Let's plug our values into the equation above.

12.2ft = HJ + 3.1ft

Subtract 3.1ft from both sides.

9.1ft = HJ

4 0
3 years ago
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