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Sunny_sXe [5.5K]
3 years ago
9

What is the answer ?​

Mathematics
2 answers:
kondaur [170]3 years ago
4 0
Lol where’s the question of
Anika [276]3 years ago
3 0

Answer: What is the question

Step-by-step explanation:

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B

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If you help me I give you branliest
Vanyuwa [196]

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m \angle MNO = \frac{1}{2} \times m MN\\\\

              = \frac{1}{2} \times 150 \\\\= 75^ \circ

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3 years ago
Solve for x: −2(x + 3) = −2x − 6<br><br> 0<br> 3<br> All real numbers<br> No solution
Alik [6]

Answer:

  x ∈ All real numbers

Step-by-step explanation:

When the distributive property is applied to the left side, the parentheses can be eliminated and the equation becomes ...

  -2x -6 = -2x -6

This is true for all possible values of x, "all real numbers".

8 0
3 years ago
Read 2 more answers
Kamal wrote the augmented matrix below to represent a system of equations.
Sergio039 [100]

Answer:

\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]

Step-by-step explanation:

Given [Missing from the question]

\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]

Required

R_2 \to -3R_2

This implies that, we form a new matrix where the second row of the new matrix is a product of -3 and the second row of the previous matrix.

So, we have:

Initial =\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]

New =\left[\begin{array}{cccc}1&0&1&|-1\\-3*1&-3*3&-3*-1&|-3*-9\\3&2&0&|-2\end{array}\right]

New =\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]

7 0
2 years ago
Read 2 more answers
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
3 years ago
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