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SIZIF
6 days ago
12

A skyscraper casts a shadow 200 feet long. If the angle of elevation of the sun is 49 , then the height of the skyscraper is ___

__.

Mathematics
2 answers:
PIT_PIT [12.4K]6 days ago
8 0
Answer: Height of the skyscraper = 230.07 feet Step-by-step explanation: In the triangle presented, AB indicates a skyscraper with a shadow extending 200 feet. If the angle of elevation of the sun is 49° In the right angle triangle ABC tanC = AB = 1.1504×200 = 230.07 feet Hence, the height of the skyscraper amounts to 230.07 feet.
zzz [12.3K]6 days ago
7 0
Given the provided data, it can be calculated using this formula: Assuming the height of the skyscraper is represented by 'y' Tan 49 = y/200 1.1504 = y/200 y = 230.8 Feet hope this assists.
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Suppose your statistics professor reports test grades as​ z-scores, and you got a score of 1.57 on an exam. ​a) Write a sentence
tester [12383]

Answer:

a) In this case, we have a z-score of 1.57, characterized as:

z =\frac{x -\mu}{\sigma}

This signifies that our score is 1.57 standard deviations above the average of all test scores.

b) P(Z

Using the normal standard distribution or Excel, we computed:

P(Z

This represents 2.275% of the dataset.

Step-by-step explanation:

Previous concepts

Normal distribution denotes a "probability distribution that is symmetric around the mean, indicating that data close to the mean occurs more frequently than data further from it".

The Z-score serves as "a statistical measurement representing a value's relation to the average (mean) of a set, calculated in terms of standard deviations away from the mean".

Solution to the problem

Part a

For this instance, we hold a z-score of 1.57, which is defined as:

z =\frac{x -\mu}{\sigma}

This shows that our score is 1.57 deviations above the overall test score average.

Part b

A z-score of z=-2 indicates that your friend's score is 2 deviations below the other test scores.

Assuming a normal distribution, we can derive the percentage:

P(Z

Using the normal standard distribution or Excel, we discovered:

P(Z

This represents 2.275% of the data.

6 0
1 month ago
Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was
tester [12383]

Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

H_0: \mu=15\\\\H_a: \mu\neq15

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

7 0
1 month ago
"The difference between five-halves of a number<br> and 17 is 48."
Inessa [12570]
Let the number be x. The equation x/5 - 17 = 48 leads to (x - 85)/5 = 48. Multiplying gives x - 85 = 240, so x = 240 + 85, resulting in x = 325.
5 0
28 days ago
A bird sanctuary needs at least 500 lbs of corn per month. The storage shed
Leona [12618]

Answer:

x≥(500-87)/40

Step-by-step explanation:

You require 500 lbs and you're starting with 87, so you must acquire at least 413 bags. Since they are sold in 40 pound bags, you divide by 40. You must round up to ensure you have enough bags to meet the 500 lbs needed.

8 0
13 days ago
A bag contains chips of which 27.5 percent are blue. A random sample of 5 chips will be selected one at a time and with replacem
lawyer [12517]

Answer:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997

Step-by-step explanation:

It is known that the mean and standard deviation of the sampling distribution of the sample proportion(\hat{p}) are represented as follows:-

\mu _{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}

, where p= Population proportion and n = sample size.

Let p denote the proportion of blue chips.

According to the information provided, we have

p= 0.275

n= 5

Thus, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{ 0.275(1- 0.275)}{5}}\\\\=0.19968725547\approx0.1997

Therefore, you will have the mean and standard deviation for the sample proportion of blue chips for samples of size 5:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997

6 0
19 days ago
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