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3 years ago

12

A tree casts a 25 foot shadow. At the same time of day, a 6 foot man standing near the tree casts a 9 foot shadow. What is the a

pproximate height of the tree to the nearest foot?

2 answers:

AysviL [449]3 years ago

7 0

So, the man is 6 feet, but casts a 9 foot shadow. So, you divide 25 by 9, and then you would find it goes into it 2.8 times, so round and get 3 if you want,or keep it. And then 12, or round to 13.

storchak [24]3 years ago

3 0

Could you solve this using similar triangles?

6/x = 9/25
cross multiply
9x = 150

x = 16.66

Assuming it can be done this way, the tree should be about 17 feet tall

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Question 2 Two trains leave the station at the same time, one heading west and the other east. The westbound train travels at 95

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Answer:

it will take 1.4 hours for the two trains to be 294 miles apart

Step-by-step explanation:

Let t be the time taken for each train

The westbound train travels at 95 miles per hour.

Speed of westbound train = 95

time = t

Distance = speed * time = 95 t

The eastbound train travels at 115 miles per hour

Speed of eastbound train = 115

time = t

Distance = speed * time = 115 t

both trains are 294 miles apart means the distance between both trains are 294 miles

So we add the distance of both trains and set it equal to 294

95t + 115t = 294

210 t =294

t = 1.4

So, it will take 1.4 hours for the two trains to be 294 miles apart

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3 years ago

2 people working at the same rate will take 6 hours to paint a room.

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Answer:4 hours

Step-by-step explanation:

6*2=12

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2 years ago

The following results come from two independent random samples taken of two populations.

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Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

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2 years ago

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Elden [556K]

Answer:

<u>Consider the following identity:</u>

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(2 + 1/2)(2² - 2*1/2 + 1/2²) =
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3 years ago

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