Answer:
5.6 days
Step-by-step explanation:
We are given;
Initial Mass; N_o = 25 g
Mass at time(t); N_t = 25/2 = 12.5 (I divide by 2 because we are dealing with half life)
k = 0.1229
Formula is given as;
N_t = N_o•e^(-kt)
Plugging in the relevant values;
12.5 = 25 × e^(-0.1229t)
e^(-0.1229t) = 12.5/25
e^(-0.1229t) = 0.5
(-0.1229t) = In 0.5
-0.1229t = -0.6931
t = -0.6931/-0.1229
t = 5.6 days
Answer:
The frequency of the note a perfect fifth below C4 is;
B- 174.42 Hz
Step-by-step explanation:
Here we note that to get the "perfect fifth" of a musical note we have to play a not that is either 1.5 above or 1.5 below the note to which we reference. Therefore to get the frequency of the note a perfect fifth below C4 which is about 261.63 Hz, we have
1.5 × Frequency of note Y = Frequency of C4
1.5 × Y = 261.63
Therefore, Y = 261.63/1.5 = 174.42 Hz.
Answer and Step-by-step explanation:
a) H o : there is no relation between job pressure and age
H 1: there is a relation between job pressure and age
b) See file attached.
c) Test statistic: For the given categorical data, we apply the x² — test Test statistic is 2.19
d) The test statistic value x² = 2.19 < 16.81, we accept the null hypothesis. Therefore, we conclude that there is no relationship between job pressure and age.
Answer:
last option
Step-by-step explanation:
We know that a₁ (or the first term) = 4 so we can rule out the second and third options. d (or the common difference) = 8 (because 4 + 8 = 12, 12 + 8 = 20 and so on) so our answer is the last option.
Ten to the <span>negative second power L over one cL is the one among the following choices that is a possible equivalence that can be used in a conversion. The correct option among all the options that are given in the question is the third option or option "C". I hope the answer helps you.</span>
