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Simora [160]
3 years ago
13

A área do círculo mede 196 π cm². Qual a medida do seu raio?

Mathematics
2 answers:
AleksandrR [38]3 years ago
8 0

Hey there!!

  • The area of a circle = πr²

The area is given as 196π

Hence,

196π = πr²

... r²= 196

... r = √196

... r = 14

<em>Hence, the radius is 14 cm. </em>

<em>Hope it helps!</em>

Rzqust [24]3 years ago
7 0

Answer:


Step-by-step explanation:

The area formula for a circle is A = pi*r^2.  Solve this for the radius (raio):

r^2 = A / pi, or r = plus or minus √(A/pi).

Here, r =  +√(A/pi) = +√( 196 pi cm^2   /   pi ) = +7.9 cm.

Check:  A = pi*r^2.  Does pi*(7.9 cm)^2 = pi*196 cm^2?   Yes.

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Answer: No she is not.

Step-by-step explanation: The correct answer is 7.25 F, because she just put them both as positives. When you have a negative it takes away from the total, so how I think of it is you subtract from the positive number, so you could just do 11.75-4.5. Hope this helped!

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B=(2a/h) - a is the answer.
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The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
cluponka [151]

Answer:

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

Step-by-step explanation:

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

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

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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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The answer is b x=4/rs
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