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3 years ago

11

5. A cube has edges 4 inches long. Susan plans to completely cover each face of the cube with square pictures that are 2 inches

long on each side. How many pictures will Susan use if
she does not overlap any of them?
12
24
20
16

1 answer:

Gnesinka [82]3 years ago

8 0

Answer:

24

Step-by-step explanation:

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5 4/5 is greater than less than or equal to 58/10?

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Deez butts, hottttie

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3 years ago

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A certain tennis player makes a successful first serve 6969% of the time. Suppose the tennis player serves 9090 times in a matc

Wewaii [24]

Answer:

a) 4.387

b) Yes, because np & npq are greater than 10.

c) = 0.017

Step-by-step explanation:

Give data:

p = 0.69

n = 90

a) a

E(X) = np = 62.1

$SD(X) = \sqrt{(np(1-p))}$

$=\sqrt{90\times 0.69(1- 0.69)}$

= 4.387

b)

np = 62.1

q = 1 - p = 1 - 0.69 = 0.31

npq = 19.251

Yes, because np & npq are greater than 10.

c.

$P(X \geq 72 ) = P(X > 71.5)$ [continuity correction]

$= P(Z> \frac{((71.5-62.1)}{ 4.387})$

= P(Z> 2.14 )

= 1 - P(Z<2.14)

= 1 - 0.983 (using table)

= 0.017

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4 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

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3 years ago

How would you do these problems i don't understand how to do them? (Attached!)

d1i1m1o1n [39]

#9 . D Feline = $1.30/lbs Kitty Kibbles = $1.40/lbs
#10.B 5:6 = 30:36

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3 years ago

Alicia uses 3/4 gallon of paint to paint her room. Becca uses for 4/5 gallon more than Alicia to paint her room. How many gallon

Maurinko [17]

Answer:

23/10 gallons

Step-by-step explanation:

Alicia uses 3/4 gallon of paint to paint her room.

Alicia use = 3/4 gallon of paint

Becca uses for 4/5 gallon more than Alicia to paint her room.

Becca uses = Alicia (3/4 gallons ) + 4/5

Becca uses = $\frac{3}{4} + \frac{4}{5}$

Total paint used together = Aliccia used + Becca used

$\frac{3}{4}+\frac{3}{4} + \frac{4}{5}$

Add like fractions 3/4 + 3/4 becomes 6/4

$\frac{6}{4} + \frac{4}{5}$

Make the denominators same

$\frac{6*5}{4*5} + \frac{4*4}{5*4}$

$\frac{30}{20} + \frac{16}{20}$

$\frac{46}{20}$

Divide top and bottom by 2

$\frac{23}{10}$ gallons

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3 years ago