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ArbitrLikvidat [17]
3 years ago
15

1/8 + 5/8?

Mathematics
1 answer:
Ronch [10]3 years ago
8 0

Answer:

6/8

4/10

9/12

9/20

9/20

Step-by-step explanation:

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In the graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded and labeled B, and the area where f(x)
mamaluj [8]

Answer:

First option:  \left \{ {{y\leq -2x + 3} \atop {y \leq x + 3}} \right.

Step-by-step explanation:

The missing graph is attached.

The equation of the line in Slope-Intercept form is:

y=mx+b

Where "m" is the slope and  "b" is the y-intercept.

We can observe that:

1. Both lines have the same y-intercept:

b=3

2. The lines are solid, then the symbol of the inequality must be \leq or \geq.

3. Since both shaded regions are below the solid lines, the symbol is:

\leq

Based on this and looking at the options given, we can conclude that the graph represents the following system of inequalities:

\left \{ {{y\leq -2x + 3} \atop {y \leq x + 3}} \right.

6 0
3 years ago
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The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims
kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is \mu=600, we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it \sigma) is

\sigma=48

Next they draw a random sample of n=70 students, and they got a mean score (denoted by \bar x) of \bar x=613

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis H_0:\bar x \geq \mu

- The alternative would be then the opposite H_0:\bar x < \mu

The test statistic for this type of test takes the form

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\

<h3>since 2.266>1.645 we  can reject the null hypothesis.</h3>
6 0
3 years ago
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Help plz i have a f in this class
Mnenie [13.5K]

Answer:

For number 3 the answer is -4

4 0
2 years ago
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Use the Venn diagram to list the elements of the set in the roster form
brilliants [131]

Answer:

Write a 75-100 word essay on the topic "Should people keep animals as pets?" No copying from other websites, I will mark your answer brainiest If I Like It!

Step-by-step explanation:

Write a 75-100 word essay on the topic "Should people keep animals as pets?" No copying from other websites, I will mark your answer brainiest If I Like It!

Write a 75-100 word essay on the topic "Should people keep animals as pets?" No copying from other websites, I will mark your answer brainiest If I Like It!

4 0
3 years ago
The equation a<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class=
andrezito [222]

Answer:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right)  = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Step-by-step explanation:

We are given the equation:

ax^2+bx+c=0

Which has roots α and β.

And we want to express (α + 1)(β + 1) in terms of <em>a</em>, <em>b</em>, and <em>c</em>.

From the quadratic formula, we know that the two solutions to our equation are:

\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Let <em>x</em>₁ = α and <em>x₂ </em>= β. Substitute:

\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)

Combine fractions:

\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)

Rewrite:

\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}

Multiply and group:

\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}

Difference of two squares:

\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}

Expand and simplify:

\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}

Distribute:

\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}

Cancel like terms:

\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}

Factor:

\displaystyle =\frac{4a(a+c-b)}{4a(a)}

Cancel. Hence:

\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Therefore:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}

4 0
3 years ago
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