Ratio of areas of similar triangles is 9 : 25.
Solution:
Given data:
Ratio of sides of two similar triangles = 3 : 5
To find the ratio of areas of the triangles:
We know that,
<em>In two triangles are similar, then the ratio of their area is equal to the square of the ratio of their sides.</em>
Ratio of areas of similar triangles is 9 : 25.
Answer:
<h3>★ <u>11/30</u> is the right answer. ★</h3>
Step-by-step explanation:
- Number of male students who got 'A' in the test is 11
- Number of female students who got 'A' in the test is 19
- Total students who got 'A' in the test is 30
- Probability that the male student got an 'A' is P(A | male) = (Number of male students who got 'A' in the test)/(Number of total students who got 'A' in the test) = <em><u>11/30</u></em>
The distributive property....u r basically multiplying the number outside of the parenthesis by everything inside the parenthesis...this gets rid of the parenthesis.
3(-4x + 8)...distribute the 3 thru the parenthesis
(3 * -4x) + (3 * 8) ...take that 3 and multiply it by every number in the parenthesis..to get rid of the parenthesis
-12x + 24 <===
4(x - 6y) =
(4 * x) - (4 * 6y) =
4x - 24y <===
6(5 - q) =
(6 * 5) - (6 * q) =
30 - 6q <===
1/2(c - 8) =
(1/2 * c) - (1/2 * 8) =
1/2c - 4 <===
-3(5 - b) =
(-3 * 5) - (-3 * b) =
-15 - (-3b) =
-15 + 3b <===
(d + 2)(-7)....re-arrange
-7(d + 2) =
(-7 * d) + (-7 * 2) =
-7d + (-14) =
-7d - 14 <===
Angle mes BCD = (mes Arc AE-mes ARC BD)/2
Plug: mes BCD = (64+20)/2 = 44° (Number C)
Answer:
1. d. Two sided t-test
2. b. 0.000193
3. a. Since p- value is very small we can conclude that the more soda the student drink the heavier they get.
4. c. Test in inconclusive; thus we cannot claim that there is a correlation between drinking soda and weight.
Step-by-step explanation:
The hypothesis testing is used to test the claim against the alternative hypothesis. The p value is determined in order to conclude the result for the test as whether to accept or reject the null hypothesis. The null hypothesis can be accepted when the p value is greater than the critical value of the t-test.
