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Kipish [7]
3 years ago
8

A rectangular park is 40 m long and 24 m wide. A path 3m wide is constructed outside the park. Find the area of the path?

Mathematics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

area of path is  area of park with path - area of rectangular park                      

area of park =l x b= 45 x 30=1350 cm2

area of park with path is =45+2.5+2.5 x 30+2.5+2.5 = 1750cm2

area of path = 1750-1350=400 cm2

Step-by-step explanation:

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C. 35/4

Step-by-step explanation:

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How many ways are there to choose a half dozen donuts from 10 varieties a)If there are no two donuts of the same variety.b)If th
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If multiples are allowed (like "10 plain, two chocolate"), then for each of 12 donuts there are 21 possibilities.

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When is -4.75 'more than' -2.25
meriva

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Sara solved the inequality -4x + &gt; 9 and graphed the solution set. She made a mistake but cannot determine where. Her work is
Lunna [17]

Answer:

step 2 is where she made her mistake

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5 0
1 year ago
In a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access
sweet-ann [11.9K]

Answer:

No, the marketing manager was not correct in his claim.

Step-by-step explanation:

We are given that in a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access in their cars.

Suppose that the marketing manager of a car manufacturer claims that the 46.6% is based only on a sample and that 46.6% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50.

<em>Let p = population proportion of all adult Americans who want car web access</em>

SO, Null Hypothesis, H_0 : p \geq 50%   {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}

Alternate Hypothesis, H_a : p < 50%  {means that the proportion of all adult Americans who want car web access is less than 0.50}

The test statistics that will be used here is<u> One-sample z proportion statistics</u>;

                T.S.  =  \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where,  \hat p = sample proportion of Americans who indicated that they were somewhat interested or very interested in having web access in their cars =  46.6%

            n = sample of Americans = 1005

So, <u><em>test statistics</em></u>  =  \frac{0.466-0.50}{\sqrt{\frac{0.466(1-0.466)}{1005} } }

                               =  -2.161

<em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the proportion of all adult Americans who want car web access is less than 0.50 which means the marketing manager was not correct in his claim.

3 0
3 years ago
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