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Y_Kistochka [10]
3 years ago
8

Can you please solve this​

Mathematics
1 answer:
Vlad [161]3 years ago
5 0

Let the length = x and the width = y

For the length of fence we have x +x + y = 48 = 2x +y = 48

Rewrite as y = 48 - 2x

For the area we have x * y  = 254

Replace y in the are equation:

x * 48 -2x = 254

Simplify the left side:

48x - 2x^2 = 254

Subtract 254 from both sides:

48x - 2x^2 - 254 = 0

Using the quadratic formula solve for x:

a = -2, b = 48 and c = -254

x = -48 + 4√17/-4

x = 16.123 m.

The length = x = 16.123 meters.

The width = 48 - (16.123*2) = 15.754 meters

Rounding to the nearest centimeter:

Length = 16.1 ( 16 meters and 1 cm)

Width = 15.8 ( 15 meters and 8 cm)

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The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
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We conclude that there is no difference in the proportion of cockroaches that died on each surface.

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We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

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                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

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