21 and 22 because 21+22=43
Answer:
There is only one distinct triangle possible, with m∠N ≈ 33°. i hope this helps :)
Step-by-step explanation:
In △MNO, m = 20, n = 14, and m∠M = 51°. How many distinct triangles can be formed given these measurements?
There are no triangles possible.
There is only one distinct triangle possible, with m∠N ≈ 33°.
There is only one distinct triangle possible, with m∠N ≈ 147°.
There are two distinct triangles possible, with m∠N ≈ 33° or m∠N ≈ 147°.
Answer:
Then it would stay at 321 million
Step-by-step explanation:
if the birth and death rate is both 1000 then nothing will change. Population wise.
Answer:
99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.
Step-by-step explanation:
Confidence interval = mean +/- margin of error (E)
mean = 76
variance = 144
sd = sqrt(variance) = sqrt(144) = 12
n = 25
degree of freedom (df) = n - 1 = 25 - 1 = 24
confidence level (C) = 99% = 0.99
significance level = 1 - C = 1 - 0.99 = 0.01 = 1%
t-value corresponding to 24 df and 1% significance level is 2.797
E = t×sd/√n = 2.797×12/√25 = 6.7128
Lower limit = mean - E = 76 - 6.7128 = 69.2872
Upper limit = mean + E = 76 + 6.7128 = 82.7128
99% confidence interval is (69.2872, 82.7128)
Answer:
P(z>1.3) = 0.9032
Step-by-step explanation:
We are given:
Mean = 5000
Standard deviation = 1000
x = 6300
P(x>6300)=?
z-score =?
z-score = x- mean/standard deviation
z-score = 6300 - 5000/1000
z- score = 1300/1000
z-score = 1.3
So, P(x>6300) = P(z>1.3)
Looking at the z-probability distribution table and finding value:
P(z>1.3) = 0.9032
So, P(z>1.3) = 0.9032
