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kolbaska11 [484]
3 years ago
7

Mr. and Mrs. McClurrie are debating on whether or not to remove a tall tree in their front yard. Mr. McClurrie is concerned that

the tree could fall during a storm and land on their home. Mrs. McClurrie disagrees. She thinks the tree poses no danger to their home and wants to keep the tree because of the wonderful shade it provides. They decide to estimate the height of the tree using special right triangles. They stood 25 feet from the base of the tree to create a 30° angle from the ground to the top of the tree. If their home is 20 feet from the base of the tree, could the McClurrie's wait a few more years before cutting down the tree or does it pose a risk now?
Mathematics
1 answer:
hichkok12 [17]3 years ago
4 0
Tanα=y/x

y=xtanα, we are told that x=25ft and α=30°

y=25tan30° ft

y≈14.43 ft  (to nearest hundredth of a foot)

Since their home is 20 feet from the base of the tree, it poses no risk right now. 
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A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco
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No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

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 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

            n = sample of cars = 600

So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

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<em>Now at 0.05 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 we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0
3 years ago
Need help with this assoon as possible
frez [133]
10(x/2 + 8) = 11, expand

10x/2 + 80 = 11

5x/2 = 11-80= -69
5x = -138
x = - 138/5




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