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julia-pushkina [17]
1 year ago
12

Factor x3 – 7x2 – 5x + 35 by grouping. What is the resulting expression?

Mathematics
2 answers:
quester [9]1 year ago
8 0
X^3-7x^2-5x+35

= x^2(x-7) - 5 (x-7)   now by definition 

=(x^2-5)(x-7)

hope this helps
nadezda [96]1 year ago
8 0

Answer: Our final factorized form will be

(x-\sqrt5)(x+\sqrt5)(x-7)

Step-by-step explanation:

Since we have given that

x^3-7x^2-5x+35

We need to factorize the expression by grouping :

x^3-7x^2-5x+35\\\\=x^2(x-7)-5(x-7)\\\\=(x^2-5)(x-7)

Since we know that

a^2-b^2=(a+b)(a-b)

so,

x^2-5=(x-\sqrt5)(x+\sqrt5)

So, our final factorized form will be

(x-\sqrt5)(x+\sqrt5)(x-7)

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A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t
denpristay [2]

Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

\widehat{Y}=Y_1+Y_2+...+Y_{200}

=200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

 S=\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486

Then, the standard error of \widehat{Y} is

\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:=83.785

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45

The sample ratio is

r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}=0.1 , this is also the estimate of the population ratio R=\frac{\overline{Y}}{\overline{X}} .

Therefore, the estimated total weight of grain in the population using ratio estimator is

\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\, lbs

4) The variance of the ratio estimator is

var(r)=\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1}   , where \mu_x=8800/200=44lbs

=\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005

Hence, the standard error of the estimate of the total population is

\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:=21.556

Hence, a 95% bound on the error of estimates is

[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]

8 0
2 years ago
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Which statement is true about the discontinuities of the function f(x)?
Tema [17]

Answer:

There are asymptotes at x = three-halves and x = negative one-third.

Step-by-step explanation:

f(x) = (x + 1)/ (6x^2 - 7x - 3)

= (x + 1)( / (6x^2 + 2x - 9x  - 3)

= (x + 1) / (2x(3x + 1) - 3(3x + 1))

= (x + 1) / (2x - 3)(3x + 1)

Now x = 3/2 and x = -1/3 both make te denominator zero so these   are both asymptotes.

6 0
1 year ago
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susan took two tests.the probability of her passing both tests is 0.6.the probability of her passing the first test is 0.8.what
OverLord2011 [107]
Formula for this is as follows:
probability of her passing both 0.6/0.8 - first test and this is a fraction. 0.6/0.8
0.6/0.8= divide 0.6 by 0.8=0.75
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1 year ago
How much heat is required to raise the temperature of
Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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1 year ago
What is the recursive formula for this arithmetic sequence? -4, 3, 10, 17, ...
MatroZZZ [7]
<span>a1 = 3
an = a (n-1) +7</span>
5 0
1 year ago
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