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3 years ago

Evaluate k^2 − 2 k + 5 , when k = − 2

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:5

2 $2x{2} -2(2)+5$

Step-by-step explanation:

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Which radical expressions are equivalent to the exponential expression below ? Check all that apply . (13 + 8) ^ (1/2) A. (13 +

Fantom [35]

Answer:

$(B)\sqrt{21}\\(C)\sqrt{13+8}$

Step-by-step explanation:

Given the expression $(13 + 8) ^{1/2}$

Using law of indices: $a^{1/2}=\sqrt{a}$

$(13 + 8) ^{1/2}=\sqrt{13+8}$

Also:13+8=21

Therefore:

$(13 + 8) ^{1/2}=21^{1/2}\\=\sqrt{21}$

8 0

3 years ago

explain why division by zero is not allowed.Don't give an exact definition from another source.Use pictures or real-life example

Roman55 [17]

Division by zero simply would work because you cannot divid something by zero, it would be the same as not dividing it at all. take a birthday cake, if you divide it by zero, u didn’t cut it at all

6 0

3 years ago

Help me find the solution set for 8x-3=2(x-1/2)

daser333 [38]

8x -3 = 2(x-1/2)
⇒ 8x -3= 2x -2*(1/2)
⇒ 8x -3 = 2x -1
⇒ 8x -2x= -1+ 3
⇒ 6x = 2
⇒ x= 2/6
⇒ x= 1/3

The final answer is x= 1/3~

6 0

3 years ago

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

Write <br> 23/25 as a percent

Firdavs [7]

3 / 25 = x / 100

<span>23/25 = 0.92 = 92/100 </span>

<span>x = 92%</span>

8 0

3 years ago

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