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لمّاح✨ SEU

لمّاح✨ SEU

قناة بسيطة

📈 نظرة تحليلية على قناة تيليجرام لمّاح✨ SEU

تُعد قناة لمّاح✨ SEU في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 11 517 مشتركاً، محتلاً المرتبة 17 202 في فئة التعليم والمرتبة 3 124 في منطقة الولايات المتحدة.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 11 517 مشتركاً.

بحسب آخر البيانات بتاريخ 25 أغسطس, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -407، وفي آخر 24 ساعة بمقدار -16، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 12.32‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 3.06‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 420 مشاهدة. وخلال اليوم الأول يجمع عادةً 353 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 4.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل قَنَاة, مَادَّة, كُلِّيَّة, مُتَطَوِّع, مَسَاء.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 26 أغسطس, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

11 517
المشتركون
-1624 ساعات
-1817 أيام
-40730 أيام
أرشيف المشاركات
3:00 To 5:00 💻 SCI101❌ MATH241❌ 5:30 To 7:30💻 SCI201 ❌ ARB260 ✖️

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Question: The mean, median and mode are same for a symmetric distribution. Answer: True

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).

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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

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Question 5 The missing F-statistic value is $\checkmark$ D. 2.0 Question 6 With 0.05 significance level, $H_a: p > 0.5$, and the value of test statistics is $z = 1.86$, then the P-value is equal to (Given that $P(z < 1.86) = 0.9686$) $\checkmark$ C. 0.0314 Question 7 A company claims that at least 65% of its customers are satisfied with their product. In a sample of 80 customers, 60 report being satisfied. To test the claim at a 0.05 significance level, the calculated test statistic for this hypothesis test is $\checkmark$ B. $z = 1.875$ Question 8 For a hypothesis test about a population mean, if the level of significance is 0.05 and the P-value is 0.50, then which of the following decision will you make: $\checkmark$ C. fail to reject $H_0$ Question 9 The calculated test statistic for a right tail test is $z = 1.0$, and the critical value of $z = 1.97$. Then, the decision would be to: $\checkmark$ B. Fail to reject $H_0$ since $z < 1.97$

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Question 5 The missing F-statistic value is $\checkmark$ D. 2.0 Question 6 With 0.05 significance level, $H_a: p > 0.5$, and the value of test statistics is $z = 1.86$, then the P-value is equal to (Given that $P(z < 1.86) = 0.9686$) $\checkmark$ C. 0.0314 Question 7 A company claims that at least 65% of its customers are satisfied with their product. In a sample of 80 customers, 60 report being satisfied. To test the claim at a 0.05 significance level, the calculated test statistic for this hypothesis test is $\checkmark$ B. $z = 1.875$ Question 8 For a hypothesis test about a population mean, if the level of significance is 0.05 and the P-value is 0.50, then which of the following decision will you make: $\checkmark$ C. fail to reject $H_0$ Question 9 The calculated test statistic for a right tail test is $z = 1.0$, and the critical value of $z = 1.97$. Then, the decision would be to: $\checkmark$ B. Fail to reject $H_0$ since $z < 1.97$