Answer:
Todd has 540540 songs in his playlist.
Vlad has 270270 songs in his playlist.
Becca has 162162 songs in her playlist.
Step-by-step explanation:
# of songs Todd has = T
# of songs Vlad has = V
# of songs Becca has = B
1) Todd has twice as many songs in his playlist as Vlad:
T=2V
2) Becca has three times as many songs in her playlist as Todd:
B=3T
3) All three people have a total of 243243 songs in their playlists:
T+V+B=243243
Because T=2V then you can substitute T as 2V in the second equation:
B=3(2V)
B=6V
Now in the final equation, you can substitute everything for V:
2V+V+6V=243243
9V=243243
V=27027
Now substitute for T:
T+1/2T+3T=243243
9/2T=243243
T=54054
Now substitute for B:
1/3B+1/6B+B=243243
3/2B=243243
B=162162
Check answers:
27027+54054+162162=243243
243243=243243
so you take -10.5+5.3+20.2=??
-10.5+5.3=-5.2
-5.2+20.2=15
Ur Answer Is 15
Did I Help??
Hope I Did!!
Answer:
4cm
Step-by-step explanation:
Given data
L=(2x+3)
W=(x-1)
P=28cm
A= L*W
A= (2x+3)*(x-1)
open bracket
A= 2x^2-2x+3x-3
collect like terms
A= 2x^2+x-3
P= 2L+2W
P= 2*(2x+3)+2(x-1)
P= 4x+6+2x-2
collect like terms
P= 6x-4
but p= 28
28= 6x-4
28-4= 6x
24= 6x
x= 24/6
x= 4cm
Hence x= 4cm
I want more points next time....Hehehee!
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<u>______________________</u></h2>
<em>Solution in attachments!</em>
Answer:
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So
X = 8.6



has a pvalue of 0.8413
X = 6.4



has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds