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Evgen [1.6K]
3 years ago
13

Find the value x ? (2x -10degree) (3x +20degree)​

Mathematics
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

If we do square on both sides we get two answers i.e. (x+12)^2=(3x+20)^2

(x^2)+24x+144=(9x^2)+120x+400

(8x^2)+96x+256=0

(x^2)+12x+32=0

(x+4)(x+8)=0

x= -4 or x= -8.

Else we have another method

x+12=3x+20

x+12–20=3x

x-8=3x,

Now square on both sides.

(x-8)^2=(3x)^2

x^2–16x+64=(9x^2)

(9x^2)-(x^2)+16x-64=0

(8x^2)+16x-64=0

(x^2)+2x-8=0

(x+4)(x-2)=0

x= -4 or x= 2.

But It Didn’t Satisfy for The Equation When

x=2 or x= -8,So The Value of x will be -4 which satisfies given question.

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(3a2 – 5ab + b2) + (–3a2 + 2b2 + 8ab) Which of the following shows the sum of the polynomials rewritten with like terms grouped
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At Simba Travel Agency, the price of a climbing trip to Mount Kilimanjaro includes an initial fee plus a constant fee per meter.
Alexus [3.1K]

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Given: At Simba Travel Agency, the price of a climbing trip to Mount Kilimanjaro includes an initial fee plus a constant fee per meter.

F(d)models the fee (in dollars) for climbing d meters.

F(d)=110+0.12d

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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.

8 0
3 years ago
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