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Sergeu [11.5K]
3 years ago
6

A fenced in rectangular region with the dimensions 4 yards by 8 yards is suddenly expanded by the same distance in each dimensio

n the resulting region is now 60 square yards in size how much distance was added to each dimension
Mathematics
1 answer:
Korolek [52]3 years ago
4 0

Answer:

The distance added to each dimension is 2 yards.

Step-by-step explanation:

The initial dimensions of the rectangular fence is 8 yards by 4 yards.

The initial area of the rectangular fence = area of a rectangle

area of a rectangle = length x width

So that,

The initial area of the fence = 8 x 4

                                              = 32

The initial area of the fence is 32 square yards.

But, with the new dimensions, area = 60 square yards.

(4 + x) x (8 + x) = 60

x^{2} + 12x + 28 = 60

x^{2} + 12x - 32 = 0

(x + 14) = 0 or (x - 2) = 0

x = -14 or x =2

Thus, x = 2

The distance added to each dimension is 2 yards.

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National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a deskt
Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, H_0 : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>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 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, <u><em>test statistics</em></u>  =  \frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }

                               =  2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis 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 doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

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Answer:

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