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Gennadij [26K]
2 years ago
14

The walk to the football field from our front door is 122

Mathematics
1 answer:
NikAS [45]2 years ago
5 0

Answer:

368 feet

Step-by-step explanation:

122 yards converted to feet is 366, and you have to add the 2 feet from before, which makes it 368 feet.

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UkoKoshka [18]
When you see the fraction 33/16, you realize that 33 divided by 16 is 2 remainder 1, so you add the 2 to the original 6, then put the remainder 1 on top of 16 ,which gives you 8 1/16
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David has scored 81 , 87 , 69 , and 82 on his previous four tests. what score does he need on his next test so that his average
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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}  } }

                               =  -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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Help please<br>h - (-2) ≥ 10​
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Answer:

answer

Step-by-step explanation:

h+2 ≥ 10

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3 years ago
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