Answer:
I believe there are 20 Neutrons in Calcium.
Explanation:
The density of the substance is obtained by dividing its mass by its volume. Density is an intensive property which means that its value does not depend on the amount of substance and will always stay the same for same conditions.
d = 84.7 g / 46.7 cm³ = 7.75 g / x cm³
The value of x is approximately 4.27 cm³.
Answer:
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
Explanation:
given data
Normal distribution
mean cholesterol level μ= 51.6 mg/dL
Standard deviation σ= 14.3 mg/dL
x= 60 mg/dL
We have to find out P(x>60)
We Know that 
therefore, 
= P(Z>0.61)
= 1 - P(Z<0.61)
= 1 - 0.7291
= 0.27
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
What's your question? is it seprete questions
Answer:
a) 20.29N
b) 19.42N
c) 15N
Explanation:
To find the magnitude of the resultant vector you can consider an axis in the middle of the vector, from which you can calculate the components of the vectors by using the angles given:
a) for 30°
![F_1=[9cos(15\°)\hat{i}+9sin(15\°)\hat{j}]N\\\\F_1=[8.69\hat{i}+2.32\hat{j}]N\\\\F_2=[12cos(15\°)\hat{i}-12sin(15\°)\hat{j}]N\\\\F_2=[11.59\hat{i}-3.10\hat{j}]N\\\\F=F_1+F_2=20.28N\hat{i}-0.78N\hat{j}\\\\|F|=\sqrt{(20.28N)^2+(0.78N)^2}=20.29N](https://tex.z-dn.net/?f=F_1%3D%5B9cos%2815%5C%C2%B0%29%5Chat%7Bi%7D%2B9sin%2815%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_1%3D%5B8.69%5Chat%7Bi%7D%2B2.32%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B12cos%2815%5C%C2%B0%29%5Chat%7Bi%7D-12sin%2815%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B11.59%5Chat%7Bi%7D-3.10%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF%3DF_1%2BF_2%3D20.28N%5Chat%7Bi%7D-0.78N%5Chat%7Bj%7D%5C%5C%5C%5C%7CF%7C%3D%5Csqrt%7B%2820.28N%29%5E2%2B%280.78N%29%5E2%7D%3D20.29N)
F = 20.29N
b) for 45°
![F_1=[9cos(22.5\°)\hat{i}+9sin(22.5\°)\hat{j}]N\\\\F_1=[8.31\hat{i}+3.44\hat{j}]N\\\\F_2=[12cos(22.5\°)\hat{i}-12sin(22.5\°)\hat{j}]N\\\\F_2=[11.08\hat{i}-4.59\hat{j}]N\\\\F=F_1+F_2=19.39N\hat{i}-1.15\hat{j}\\\\|F|=\sqrt{(19.39N)^2+(1.15N)^2}=19.42N](https://tex.z-dn.net/?f=F_1%3D%5B9cos%2822.5%5C%C2%B0%29%5Chat%7Bi%7D%2B9sin%2822.5%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_1%3D%5B8.31%5Chat%7Bi%7D%2B3.44%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B12cos%2822.5%5C%C2%B0%29%5Chat%7Bi%7D-12sin%2822.5%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B11.08%5Chat%7Bi%7D-4.59%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF%3DF_1%2BF_2%3D19.39N%5Chat%7Bi%7D-1.15%5Chat%7Bj%7D%5C%5C%5C%5C%7CF%7C%3D%5Csqrt%7B%2819.39N%29%5E2%2B%281.15N%29%5E2%7D%3D19.42N)
F= 19.42N
c) for 90°
for this case you can consider that the direction of both vectors are the y and x axis of the Cartesian plane:

F=15N