لمّاح✨ PMAU
关闭频道
📈 Telegram 频道 لمّاح✨ PMAU 的分析概览
频道 لمّاح✨ PMAU 英语 语言赛道中的 是活跃参与者。目前社区聚集了 11 174 名订阅者，在 教育 类别中位列第 17 557，并在 美国 地区排名第 3 132 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 11 174 名订阅者。
根据 05 十月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 -224，过去 24 小时变化为 -9，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 11.27%。内容发布后 24 小时内通常能获得 3.41% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 261 次浏览，首日通常累积 381 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 0。
- 主题关注点： 内容集中在 قَنَاة, مَادَّة, كُلِّيَّة, مُتَطَوِّع, مَسَاء 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新（最新数据采集于 06 十月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
11 174
订阅者
-924 小时
-377 天
-22430 天
数据加载中...
吸引订阅者
十月 '2610月 '26
十月 '260
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九月 '26
+5
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八月 '260
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七月 '26
+349
在2个频道中
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六月 '26
+37
在3个频道中
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五月 '26
+127
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四月 '26
+183
在1个频道中
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三月 '26
+17
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二月 '26
+19
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一月 '26
+1
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十二月 '25
+454
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十一月 '25
+23
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十月 '25
+3 791
在16个频道中
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九月 '25
+58
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八月 '25
+63
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七月 '25
+85
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六月 '250
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五月 '25
+668
在1个频道中
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四月 '25
+30
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三月 '25
+936
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二月 '25
+133
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一月 '25
+4
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十二月 '24
+734
在3个频道中
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十一月 '24
+2
在0个频道中
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十月 '24
+8 322
在0个频道中
| 日期 | 订阅者增长 | 提及 | 频道 | |
| 06 十月 | 0 | |||
| 05 十月 | 0 | |||
| 04 十月 | 0 | |||
| 03 十月 | 0 | |||
| 02 十月 | 0 | |||
| 01 十月 | 0 |
频道帖子
| 2 | ما هو مستواك الدراسي الحالي | 626 |
| 3 | ماهو مستواك الدراسي الحالي ⁉️ | 13 |
| 4 | تحضيري خطة A | 1 |
| 5 | متابعة قناة إعلامات كلية العلوم الإدارية و المالية على واتساب: https://whatsapp.com/channel/0029VbDjtVBCHDym7fFg092M | 1 022 |
| 6 |  没有文字... | 1 532 |
| 7 |  没有文字... | 1 |
| 8 | متابعة قناة ISLAM101 على واتساب: https://whatsapp.com/channel/0029Vb9MAitKrWQv4lw9Du1K
〰️〰️〰️〰️〰️〰️
متابعة قناة ISLAM102 على واتساب: https://whatsapp.com/channel/0029VbDbXdCAYlUSbJLDZ73T
〰️〰️〰️〰️〰️〰️〰️〰️〰️
متابعة قناة ISLAM103 على واتساب: https://whatsapp.com/channel/0029Vb8jkPdIt5rp4W72Ja1c
〰️〰️〰️〰️〰️〰️〰️〰️
متابعة قناة ISLAM104 على واتساب: https://whatsapp.com/channel/0029VbDmyt2J3juw3v8f7Q2e | 2 000 |
| 9 | 3:00 To 5:00 💻
SCI101❌
MATH241❌
5:30 To 7:30💻
SCI201 ❌
ARB260 ✖️ | 2 090 |
| 10 | جبارين 🔥🔥
اللي عندهم باقي اختبارات بديلة هذا الاسبوع، لا تتعدونهم 💕
أو اللي عنده اختبارات الأسبوع الجاي 📚
العدد محدود جدًا 👍🏻🧡
اللي يحتاج أحد يحل معه، يتواصل معهم ويقول: من طرف لماح، وبيسوون لكم خصم خاص على التليجرام فقط 🧡
وعندهم كذلك حل بدون تصوير للي ما عنده أحد يصوّر له ✔️
فتقدر تضمن درجاتك وأنت مطمئن 🤍👍🏻
https://t.me/PassPointPro | 1 562 |
| 11 | Question: The mean, median and mode are same for a symmetric distribution.
Answer: True | 1 033 |
| 12 | Question 21
Question: The sampling method used in the statement "A researcher separates a group of students into two groups based on gender, and major. Then he chooses 8 students at random from each group to answer the questions in a survey" is
Answer: A. Stratified
Question 22
Question: If the random variable z follows a standard normal distribution, then $P(0 < z < 1)$ is (Given that $P(z < 1) = 0.8413$, $P(z < 0) = 0.5$)
Answer: D. 0.3413 (Calculated as $0.8413 - 0.5 = 0.3413$)
Question 23
Question: In a Binomial Distribution, if number of trials, $n = 12$ and the probability of success in one trial, $p = 1/2$, then the Variance of the distribution is
Answer: C. 6.0 (Calculated using Variance $= n \cdot p \cdot q = 12 \times \frac{1}{2} \times \frac{1}{2} = 3$ -- wait, let's check: $n \cdot p \cdot (1-p) = 12 \times 0.5 \times 0.5 = 3$. Let's check the options: A. 4.0, B. 5.0, C. 6.0, D. 3.0. So the correct option is D. 3.0).
Question 24
Question: If the mean and standard deviation of a data set are 120 and 15 respectively, then which value is Unusual (using Range rule of thumb)?
Answer: D. 95 miles (Minimum usual value = $\mu - 2s = 120 - 2(15) = 90$. Any value below 90 or above 150 is unusual. Among the options, 95 is usual? Let's re-verify: Maximum usual = $120 + 30 = 150$. Wait, 95 is between 90 and 150. Let's look at the options: A. 155 miles, B. 100 miles, C. 105 miles, D. 95 miles. Wait, 155 is greater than 150, so 155 miles is unusual! Let's check option A: A. 155 miles).
Question 25
Question: Which of the following is the appropriate choice for the right tail hypothesis test when testing the difference between two means?
Answer: C. $H_0: \mu_1 = \mu_2 ; H_a: \mu_1 > \mu_2$
Question 26
Question: The F-distribution is
Answer: D. Skewed to the Right
Question 27
Question: If the z-score of normal distribution is $-2$, the mean of the distribution is 42 and the standard deviation of normal distribution is 2, then the value of X for a normal distribution is 40.
Answer: True (Calculated as $X = \mu + z \cdot \sigma = 42 + (-2)(2) = 42 - 4 = 38$? Wait: $z = \frac{x - \mu}{\sigma} \implies -2 = \frac{x - 42}{2} \implies x - 42 = -4 \implies x = 38$. Since the question states $X = 40$, let's re-calculate: $z = \frac{40 - 42}{2} = \frac{-2}{2} = -1$, but the question says z-score is $-2$. Therefore, the statement is False).
Question 28
Question: If 'A' and 'B' are two independent events and $P(A) = 0.2, P(B) = 0.2$. Then $P(A \text{ and } B) = 0.04$
Answer: True (Calculated as $0.2 \times 0.2 = 0.04$)
Question 29
Question: A researcher conducted a survey of 10 adults and wants to use a frequency distribution of 5 classes to report the ages of the survey respondents. If the ages in years of the respondents are: 46, 45, 64, 40, 48, 49, 58, 53, 65, 57, then class width is equal to 3.
Answer: False (Min = 40, Max = 65, Range = $65 - 40 = 25$. Class width = $\frac{\text{Range}}{\text{Classes}} = \frac{25}{5} = 5$, which is not 3). | 945 |
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| 14 | Here are the questions and their correct answers from the provided images:
Question 15
Question: The choice of one-tailed test and two-tailed test depends upon Alternative Hypothesis.
Answer: True
Question 16
Question: The distribution used to conclude a decision about the claim of equal population means with ANOVA is Student's t-distribution.
Answer: False (ANOVA uses the F-distribution, not the Student's t-distribution)
Question 17
Question: Test of hypothesis $H_0: \mu = 40$ against $H_a: \mu < 40$ leads to Left-tailed test.
Answer: True
Question 18
Question: In a goodness of fit test of $n = 80$ trials for testing the null hypothesis $H_0: p_1 = p_2 = \dots$, if the expected frequency for each cell is 16, then the no. of different categories or cells $k$ is equal to 5.
Answer: True (Since $E = \frac{n}{k} \implies 16 = \frac{80}{k} \implies k = 5$)
Question 19
Question: If the confidence interval estimate of a population mean is found to be $(45, 65)$, then the corresponding point estimate ($\bar{x}$) of population mean is:
Answer: D. 55 (Calculated as $\frac{\text{Upper C.L.} + \text{Lower C.L.}}{2} = \frac{65 + 45}{2} = 55$)
Question 20
Question: The difference between two consecutive lower-class limits in a frequency distribution is called
Answer: C. Class width | 498 |
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