Answer:
2500÷$3.21 =778.81
just round to the nearest whole number
A. is the answer.
the 2 triangls have AD = AC
and AB as a common side. therefore the triangls are congruent.
Answer:
C. with 3000 successes of 5000 cases sample
Step-by-step explanation:
Given that we need to test if the proportion of success is greater than 0.5.
From the given options, we can see that they all have the same proportion which equals to;
Proportion p = 30/50 = 600/1000 = 0.6
p = 0.6
But we can notice that the number of samples in each case is different.
Test statistic z score can be calculated with the formula below;
z = (p^−po)/√{po(1−po)/n}
Where,
z= Test statistics
n = Sample size
po = Null hypothesized value
p^ = Observed proportion
Since all other variables are the same for all the cases except sample size, from the formula for the test statistics we can see that the higher the value of sample size (n) the higher the test statistics (z) and the highest z gives the strongest evidence for the alternative hypothesis. So the option with the highest sample size gives the strongest evidence for the alternative hypothesis.
Therefore, option C with sample size 5000 and proportion 0.6 has the highest sample size. Hence, option C gives the strongest evidence for the alternative hypothesis
In this case since 067 is less than 180, 180 will be added to it to find the bearing of b from a.
067+180=247
the bearing of b from a is 247
or
subtract 067 from 360
360-067=293
I am not sure of the right answer but I hope this helps.
Answer:
M = 1/0.000121 = 8264.5 years
Step-by-step explanation:
M = − k ∫∞₀ teᵏᵗdt
To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt
∫udv = uv - ∫ vdu
u = t
du/dt = 1
du = dt
∫ dv = ∫ eᵏᵗdt
v = eᵏᵗ/k
∫udv = ∫ teᵏᵗdt
uv = teᵏᵗ/k
∫ vdu = eᵏᵗ/k
∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k
But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²
∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²
The rest of the calculation is done on paper in the image attached to this question
