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Delicious77 [7]
3 years ago
12

In your notebook, set up the following addition using a vertical format and find the sum of the given polynomials. a - b + c, b

- c + d, and c - d + e
A) a + c + e

B) a + 2b + 3c + 2d + e

C) b + d + e

D) a - c - e
Mathematics
2 answers:
liq [111]3 years ago
5 0
Your answer would be A) a+c+e

a-b+c
b-c+d
c-d+e
+
——————
a+c+e
Alexus [3.1K]3 years ago
3 0
The answer is b I remember having this question
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Y-1=5/6(x-4) in standard form
scoray [572]
           y - 1 = ⁵/₆(x - 4)
           y - 1 = ⁵/₆(x) - ⁵/₆(4)
           y - 1 = ⁵/₆x - 3¹/₃
             + 1         + 1
                 y = ⁵/₆x - 2¹/₃
        ⁻⁵/₆x + y = ⁵/₆x - ⁵/₆x - 2¹/₃
   -6(⁻⁵/₆x + y) = -6(-2¹/₃)
-6(⁻⁵/₆x) - 6(y) = 14
         5x - 6y = 14
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3 years ago
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I need help please I’m struggling
Aloiza [94]

Answer:

It is first answer

Step-by-step explanation:

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3 years ago
Which graphed matches the equation y+6=3/4(x+4)
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Answer:

Well, this is a bit difficult to answer considering there's no provided pictures or option images of graphs, but this equation graphed looks like:

Step-by-step explanation:

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3 years ago
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Evaluate:<br><br> 16 - 12 + 13.5 - - 14
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31.5

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2 years ago
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A researcher studying the sleep habits of teens will select a random sample of n teens from the population to survey. The resear
kati45 [8]

Answer:

D. The center remains constant, and the area in the tails of the distribution decreases.

Step-by-step explanation:

Hello!

Be it two independent random variables, X~N(μ;σ²) and U~Xₙ², the variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:

t= \frac{(X-Mu)/Sigma}{\sqrt{U/n} }

As a consequence of this, the structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.

It has a mean and variance:

E(t)= 0 for n > 1

V(t)= \frac{n}{n - 2} n > 2

As you can see the variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.

When ↑n ⇒ ↑V(t) ⇒ The area under the tails increases.

When ↓n ⇒ ↓V(t) ⇒ The area under the tails decreases.

In this example, the degrees of freedom of the distribution decreased from 40 to 20, then the variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.

The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.

I hope this helps!

6 0
3 years ago
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