<span>1.6625 x 10−21</span>
<span>mass of one helium atom x 250 </span>
Express 250 as 2.50.
(6.65 x 10−24)(2.50 x 102)
16.625 x 10<span>−22</span>
Answer:
d. 9 integers
Step-by-step explanation:
Given that two numbers are said to be 'relatively prime' if their greatest common factor is 1.
We are to find relatively prime with 28 which are greater than 10 and less than 30
We have 11,12,13...29 satisfying the criteria greater than 10 and less than 30
To be relatively prime with 28, common factors should be only 1.
28 = 2x2x 7. Hence the numbers which do not have factors as 2 or 7 will be relatively prime. Remove all the even numbers from the list.
We have 11,13,15....29.
Of these 21 is the multiple of 7 so remove that.
Thus we have now 11,13,15,17,19,23,25,27,29
9 integers
d. 9 integers
Answer:
y=3x+2
Step-by-step explanation:
The change is 6 because the pouts you lose depends on how many times you leave the shoes at home
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
