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Naya [18.7K]
3 years ago
7

What is the value of F ?

Mathematics
2 answers:
fgiga [73]3 years ago
4 0

Step-by-step explanation:

110+ f =180

f = 180-110

f=70

Kruka [31]3 years ago
4 0

Answer:

here is your answer......

Step-by-step explanation:

f plus 110=180 ( because being straight line)

or; f= 180-110

or; f=70

therefore,f=70

the value of f is 70 degree

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Brilliant_brown [7]
The  answer is a 32-gon
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The length of a rectangular floor is two ft more than its width. If the area of the floor is 63 ft, find the length and the widt
NemiM [27]

Answer:

width = 7 and length = 9

Step-by-step explanation:

Let's assume width = w, and length = 2 + w

So area of a rectangle = width * length

63 ft² = w (2 + w)

63 = 2w + w²

w² + 2w - 63 = 0

Use the quadratic formula to solve this

here, a = 1, b = 2 and c = -63

w = 7 or -9

The dimensions of a rectangle cannot be negative. So we take the dimension, w = 7.

Width = 7 and Length = 7+2 = 9

3 0
2 years ago
How do you solve 125 - (5-9)^3÷2
hoa [83]
Solve what is in parenthesis first (5-9) = -4

Solve exponents next -4^3 = -64

Divide next
- 64 \div 2
= -32

Lastly, subtract and resolve the negatives which become a positive number 125 - (-32) = 125 + 32 = 157
8 0
3 years ago
I need help with Tuesday hw
Andre45 [30]
1) X^5Y/X^5=  X^0Y= Y  (Anything raise to power of 0 is 1)
2)(3mn)^1= 3mn
3)m^-4xM^5= M^-4+5= M
4) 5^-4x5^4= 5^-4+4= 5^0= 1
5) nm^3/n^3m= n^1-3m^3-1= n^-2m^2= m^2/n^2
6) (-2)^0= 1

therefore number 4 and 6 has the value of 1
3 0
3 years ago
if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?
Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • There are 9 rolls, hence n = 9.
  • Of the six sides, 2 are 3 or 4, hence p = \frac{2}{6} = 0.3333

The desired probability is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1)

Then

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026

P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0
3 years ago
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