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3 years ago

I need to know the answer

Mathematics

2 answers:

AlexFokin [52]3 years ago

4 0

Answer:

k = - 3

Step-by-step explanation:

$3k + 2k - 6 = 8k + 3 \\ 5k - 6 = 8k + 3 \\ 5k - 8k = 6 + 3 \\ - 3k = 9 \\ k = \frac{9}{ - 3} \\ \huge \red{ \boxed{ k = - 3}}$

makkiz [27]3 years ago

3 0

Step-by-step explanation:

3k + 2k - 6 = 8k + 3

8k - 3k - 2k = - 6 - 3

8k - 5k = -9

3k = - 9

k = - 9/ 3

K = - 3

Hope it will help you :)

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Please help me it's due tomorrow

vitfil [10]

The graph represents p(x) = |x|

To graph r(x) and q(x), pick any values of x to graph (I'd pick the numbers on the x- scales, -6, -4, -2, 0, 2, 4, and 6)

As for explaining them, I'm sure something like "multiply the absolute value of x by -1/2" for describing q(x) is an acceptable answer.

6 0

3 years ago

Is two over five irrational numbers

9966 [12]

Answer:

No, It is not irrational number.

Step-by-step explanation:

2/5 is a rational number. A rational number is a number of the form q/q where p and q are integers and q is not equal to 0.

4 0

3 years ago

Mia found the area of a polygon The area is 32 square cm.

solniwko [45]

Answer:

Only the isosceles trapezoid has an area of 32 cm².

Step-by-step explanation:

Let's calculate the area of each polygon.

For the two triangles we have:

$A_{t} = \frac{bh}{2} = \frac{2 cm*4 cm}{2} = 4 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle we have:

$A_{r} = bh = 6 cm*4 cm = 24 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle trapezoid we have:

$A_{rt} = A_{t} + A_{r} = (4 + 24) cm^{2} = 28 cm^{2}$

So, this polygon does not have an area of 32 cm².

Finally, for the isosceles trapezoid:

$A_{it} = A_{t}*2 + A_{r} = (4*2 + 24) cm^{2} = 32 cm^{2}$

This polygon does have an area of 32 cm².

Therefore, only the isosceles trapezoid has an area of 32 cm².

I hope it helps you!

8 0

3 years ago

Can you please help me with 3 and 4

Otrada [13]

Number 3 is 1:3 but I don't know what 4 is I can't see the whole problem

3 0

2 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago