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Svetllana [295]
2 years ago
6

Marta said 50 + one-half m = 120 is an equation, while Maribel argued that it is an expression. Determine who is correct and exp

lain why.
Mathematics
2 answers:
Alina [70]2 years ago
4 0

Sample Response: Marta is correct. This is an equation. The statement includes an equal sign. Equations include an equal sign, expressions do not include an equal sign.

In-s [12.5K]2 years ago
3 0

Answer:

Marta

Step-by-step explanation:

We are given that

50+\frac{1}{2}m=120

We have to find who is correct and explain.

We know that

Equation: It is mathematical sentence that combines  numbers and variables which shows the equality of two expressions.

Example:

x+y=3

Expression: It is mathematical phrase that combines  numbers and variables

using mathematical operators.It can be evaluated but not solved.

Example:

x+y

By definitions

50+\frac{1}{2}m=120

It is in equation form.

Therefore, Marta is correct.

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(c - b)² + a²


a = -5

b = -2

c = -4


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(-4 + 2)² + (-5)²

(-2)² + (-5)²

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A commuter must pass through 5 traffic lights on her way to work and will have to stop at each one that is red. She estimates th
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Answer:

(a) The mean or expected value of <em>X </em>is 2.2.

(b) The standard deviation of <em>X</em> is 1.3.

Step-by-step explanation:

Let <em>X</em> = number of times the traffic light is red when a commuter passes through the traffic lights.

The probability distribution of <em>X</em> id provided.

The formula to compute the mean or expected value of <em>X </em>is:

\mu=E(X)=\sum x.P(X=x)

The formula to compute the standard deviation of <em>X </em>is:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}

The formula of E (X²) is:

E(X^{2})=\sum x^{2}.P(X=x)

(a)

Compute the expected value of <em>X</em> as follows:

E(X)=\sum x.P(X=x)\\=(0\times0.06)+(1\times0.25)+(2\times0.35)+(3\times0.15)+(4\times0.13)+(5\times0.06)\\=2.22\\\approx2.2

Thus, the mean or expected value of <em>X </em>is 2.2.

(b)

Compute the value of E (X²) as follows:

E(X^{2})=\sum x^{2}.P(X=x)\\=(0^{2}\times0.06)+(1^{2}\times0.25)+(2^{2}\times0.35)+(3^{2}\times0.15)+(4^{2}\times0.13)+(5^{2}\times0.06)\\=6.58

Compute the standard deviation of <em>X</em> as follows:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}\\=\sqrt{6.58-(2.22)^{2}}\\=\sqrt{1.6516}\\=1.285\\\approx1.3

Thus, the standard deviation of <em>X</em> is 1.3.

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