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Aleksandr-060686 [28]
3 years ago
10

Bryce has 135 feet of rope. He wants to cut it into pieces that are each 9 feet long. How many pieces of rope will Bryce have?

Mathematics
2 answers:
GarryVolchara [31]3 years ago
8 0
He will have 15 pieces of rope. hope this help c:

jeka57 [31]3 years ago
8 0
He will have 15 pieces of rope because if you divide 135 by 19 you get 15! Hope this helped! tell me if u need help on anything else
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Evgen [1.6K]

Answer:

n = -1

Step-by-step explanation:

Guess your question is to find n

Given:

-6 + 3n = -9

-6 + 3n = -9

Add 6 to both sides

-6 + 6 + 3n = -9 + 6

3n = -3

Divide both sides by 3

n = -3 / 3

= -1

n = -1

Check:

-6 + 3n = -9

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3 years ago
5(8n+ 7) < 5-5(n+3)
svet-max [94.6K]

Answer:

n<-1

Step-by-step explanation:

Distribute 5 and -5 into each parenthesis respectively: 40n+35<5-5n-15

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2 years ago
Can some help me plz​
Stels [109]

Answer:

\sqrt{2}

Step-by-step explanation:

Apply the 30-60-90 formula

side between 60 and 90 is x

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If x\sqrt{3}=\sqrt{6} then x= \sqrt{2}

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3 years ago
A box contains $8.00 in nickels, dimes, and quarters. There are three times as many nickels as quarters, and the number of dimes
IgorLugansk [536]

Answer:

There are 12 quarters, 36 nickels and 32 dimes.

Step-by-step explanation:

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2 years ago
The probability density function (p.d.f.) of a continuous random variable XX is defined to be: f(x)={x6+k for 0
koban [17]

The probability density function (p.d.f.) of a continuous random variable X is defined to be:

f(x)= x/6+k for 0<x<2.50 otherwise,

for some constant k.

For these problems, please ensure your answers are accurate to within 3 decimals.

Part a) Find the value of k that makes the above function a proper p.d.f.

Part b) Hence find P(0.5<X<1).

Answer:

a. k = 0.192

b. P(0.5<X<1) = 1

Step-by-step explanation:

Given

f(x) = x/6 + k for 0 x < 2.50

To find the value of k that makes the above function a proper p.d.f.

∫ f(x) dx must be equal to 1

∫ f(x) dx = 1

Substitute f(x) = x/6 + k

∫ x/6 + k dx {0,2.50}

Integrate with respect to x

x²/(6*2) + kx {0,2.50} = 1

x²/12 + kx {0,2.50} = 1

(2.50²/12 + 2.50k) - (0²/12 - 0*k) = 1

6.25/12 + 2.50k = 1

Collect like terms

2.50k = 1 - 6.25/12

2.50k = (12 - 6.25)/12

2.50k = 5.75/12

Divide through by 2.50

2.50k/2.50 = 5.75/12 * 1/2.50

k = 2.3/12

k = 23/120

k = 0.192 --- Approx to 3 decimal places

So f(x) = x/6 + 23/120 for 0 x < 2.50

b.

Find P(0.5<X<1)

Given that f(x) = x/6 + 23/120 for 0 x < 2.50

P(0.5<X<1) = ∫x/6 + 23/120 {0,2.50}

P(0.5<X<1) = x²/12+ 23x/120 {0,2.50}

P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0²/12+ 23*0/120)

P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0)

P(0.5<X<1) = (6.25/12+ 57.5/120) - (0)

P(0.5<X<1) = (6.25/12+ 5.75/12)

P(0.5<X<1) = (6.25 + 5.75)/12

P(0.5<X<1) = 12/12

P(0.5<X<1) = 1

5 0
3 years ago
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