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Alinara [238K]
3 years ago
9

WILL GIVE BRAINLEIST : A party hat is in the shape of a cone. What is the volume of the hat in cubic centimeters?​

Mathematics
1 answer:
Brut [27]3 years ago
7 0
36xpi or 113.1 cm^3
hope that helped!! :)
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Rational

Step-by-step explanation:

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I think the answer for this question is C

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Please find the length of the diagonal, thanks. Will give brainliest
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Answer is 12.73
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Mailyn loves cookies. She can eat 32 in an hour. Her brother Juan, however, needs 3 hours to eat the same amount of cookies. How
bezimeni [28]

we know that

1) Mailyn can eat 32 cookies in an hour

1 hour  is equal to 60 minutes

<u>Find how cookies eats Mailyn in one minute</u>

by proportion

\frac{32}{60} \frac{cookies}{minutes}= \frac{x}{1} \frac{cookie}{minutes} \\\\ x=32/60\\x=0.53 \ cookies

2) Juan needs 3 hours to eat 32 cookies

<u>Find how cookies eats Juan in one minute</u>

by proportion

\frac{32}{180} \frac{cookies}{minutes}= \frac{x}{1} \frac{cookie}{minutes} \\\\x=32/180 \\x=0.18\ cookies

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we know that

Maylin and Juan eats in one minute together ---------> (0.53+0.18)=0.71 cookies

so

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\frac{0.71}{1} \frac{cookies}{minutes}= \frac{40}{x} \frac{cookie}{minutes} \\\\0.71*x=40\\x=56.34 \ minutes

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56.34\ minutes

4 0
3 years ago
You collect baseball cards and buy sealed packs from the grocery store and clearly have no idea which cards are inside (nor if t
ololo11 [35]

Answer:

(a) 0.9607

(c) 0.04

(d) 0.8184

(e) 0.074

(f) 0.0853/e^0.8;

750 cards

Step-by-step explanation:

Let the probability of success of a valuable card be p

p= 1/250 = 0.004,

q = 1-p = (249/250)

(a) There are 10 cards in one pack

The probability that there are no valuable card is given by:

Pr(X= 0) = 10C0 × (1/250)^0 ×(249/250)^10

10C0 = 10 combination 0= 10!/10!0!

=1

Pr (X= 0) = (249/250)^10

Pr (X=0) = 0.9607

(c) Expected number of valuable card is given by:

E(X) = np , n= 10, p= 1/250

E(X) = 10 × 1/250 = 1/25 = 0.04

(d) 5 packages means 5 × 10= 50 cards

Pr(X=0) = 50C0 × (1/250)^0 ×(249/250)^50

50C0 = 50!/50! ×0!=1

Pr (X=0) = (249/250)^50

Pr (X= 0) =(0.996)^50

Pr (X=0) = 0.8184

(e) For this case we use the Binomial Distribution

2 packages 2 × 10 = 20 cards

n = 20 , x = 1

We use:

P(X=1) = 20C1 × (1/250)^ 1 × (249/250)^19

Pr (X=1) = 20C1 × (1/250)^1 × (249/250)^19

Pr (X=1) = 20 × (1/250)¹ × (249/250)^19

Pr (X= 1 )= 0.074

(f) 20 packages = 200 cards

For this we apply Poisson distribution

P(X=3) = (e ^-h × h ^x) / x!

Where h = np = 200/250 = 0.8

P(X=3) = e^-0.8 × (0.8)^3 / 3!

P(X=3) = 0.991 × 0.512/6

P(X=3) = 0.0846

3 = n p

n = 3/p = 3 /0.04

n = 750 cards

8 0
4 years ago
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