Is anyone able to see if there is a counterexample for the following argument in logic?

Add: (3x2 - 5x + 6) + (9 - 8x - 4x2) A) 12x2 - 13x + 2 B) 7x2 + 3x - 3 C) 7x2 - 13x + 15 D) -x2 - 13x + 15

kati45 [8]

Answer:

D

Step-by-step explanation:

So we have the expression:

$(3x^2-5x+6)+(9-8x-4x^2)$

Combine like terms:

$=(3x^2-4x^2)+(-5x-8x)+(6+9)$

Add or subtract:

$=(-1x^2)+(-13x)+(15)$

Simplify:

$=-x^2-13x+15$

Our answer is D

4 0

3 years ago

In the 7th grade, 150 of the students are attending the dance. If this is 62.5% of the entire grade, how many 7th graders are th

JulijaS [17]

Answer:

244 students.

Step-by-step explanation:

Okay, so. We know 150 people of the 7th grade is coming and 62.5% is the entire of the 7th graders coming. We would multiply.

150 * 62.5% = 93.75

We would round it up. It is over 5, so it would be 94 students. Then we would add

150 + 94 = 244

There are 244 students in 7th grade.

3 0

3 years ago

Write the number as hundredths in fraction form and decimal form. 9/10

Agata [3.3K]

0.09

hope this helped!!

3 0

3 years ago

What two numbers add up to -13 and multiply to 36?

vodka [1.7K]

-4 and -9 is the correct answer

6 0

3 years ago

Read 2 more answers

26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0

2 years ago