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3 years ago

Please help me i need this answered before midnight tonight.

2 answers:

8 0

A. The length is 6 mm greater than the width, so width added by 6 mm is length...

w + 6 = l

b. This question is a bit vague, but here's what I think:

The perimeter of a rectangle can be found by multiplying the length and width each by 2 and adding them together:

2l + 2w = perimeter

We know from the question that the perimeter of the rectangle is greater than 44mm, so:
2l + 2w > 44

If you use l = w + 6 as well:

2(w + 6) + 2w > 44

2w+ 12 + 2w > 44

Answer: 4w+ 12 > 44

c. Basically simplifying what we got in b:
4w+ 12 > 44
4w > 32
w > 8
Answer: w > 8

Paul [167]3 years ago

6 0

Bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

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What is the solution of √x-5 = x-11

PIT_PIT [208]

Hey!

To solve x in this equation we must first add five to both sides to get $\sqrt{x}$ on its own.

Original Equation :
$\sqrt{x} - 5 = x - 11$

New Equation {Added 5 to Both Sides} :
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Now we must square both sides of the equation.

Old Equation :
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New Equation {Changed by Squaring Both Sides} :
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And now we must solve the new equation.

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Step 2 - Subtract x from both sides
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Step 3 - Simplify
$x^{2} -13x+36=0$

Now we need to solve the rest of the equation using the quadratic formula.

$\frac{-(-13)+ \sqrt{(-13) ^{2}-4*1*36 } }{2*1}$

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$\frac{13+ \sqrt{25}}{2}$

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9

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3 years ago

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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.

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3 years ago