Answer:
65
Step-by-step explanation:
remember a TRIANGLE always have a 180 degrees total angle,
so 45+70=115
180-115=65
Answer:
Han bebido la misma cantidad.
Step-by-step explanation:
De la pregunta,
Frederico = 2/5 vaso de leche
María = 4/10 vaso de leche
Simplificando el gozo de la leche para María
= 2/5
Por tanto, podemos decir que:
Sí, han bebido la misma cantidad.
Answer: provided in the explanation segment
Step-by-step explanation:
(a). from the question, we can see that since that б is known, we can use standard normal, z.
we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?
⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;
x ± z * б / √m
which is
3.15 ± 1.28 * 0.32/√10
= 3.15 ± 0.1295 = 3.0205 or 3.2795
(b). normal distribution of weight (c) б is known
(c). option (a) and (e) are correct
(d). from the question, let sample size be given as S
this gives';
1.28 * 0.32/√S = 0.15
√S = (1.28 * 0.32) / 0.15 = 2.73
S = 7.4529
cheers i hope this helps
According to the Central Limit Theorem, the distribution of the sample means is approximately normal, with the mean equal to the population mean (1.4 flaws per square yard) and standard deviation given by:
![\frac{\sigma}{ \sqrt{n} }=\frac{1.2}{ \sqrt{178} }=0.09](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%7D%7B%20%5Csqrt%7Bn%7D%20%7D%3D%5Cfrac%7B1.2%7D%7B%20%5Csqrt%7B178%7D%20%7D%3D0.09)
The z-score for 1.5 flaws per square yard is:
![z=\frac{1.5-1.4}{0.09}=1.11](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B1.5-1.4%7D%7B0.09%7D%3D1.11)
The cumulative probability for a z-score of 1.11 is 0.8665. Therefore the probability that the mean number of flaws exceeds 1.5 per square yard is
1 - 0.8665 = 0.1335.