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storchak [24]
2 years ago
14

PLEASE HELP!! I got until monday to finish this!!

Mathematics
2 answers:
uysha [10]2 years ago
6 0
Bc and gh respectively
erica [24]2 years ago
5 0
1.would be AEH.
2.would be ADH.
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Jeff saved $500 from his summer job so he would have spending money during the school year. He withdraws $12 from his account ea
Alex Ar [27]
Jeff Lucy Goethe licky
4 0
3 years ago
How do you solve 4b^2+8b+-12=0. Also find the parabole.?
Likurg_2 [28]
First you divide everything by 4 which will become b^2+2b-3=0
Then you can factor this which is
(b-1)(b+3)=0
B=1
B=-3 ... I think that's how you do it
4 0
3 years ago
Apply the distributive property and the greatest common factor to write an equivalent expression.
amid [387]
Here, Your Original expression is 60x - 24.

Take 12 as common, which is GCD of 24 & 60, 

Therefore, It would be: 

60x - 24
= 12 * 5x - 12 * 2
= 12 (5x - 2)

In short, your Answer would be 12(5x - 2)

Hope this helps!

Photon
5 0
3 years ago
Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.
Georgia [21]
Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\  \\ 
z-score =2.326

So, 90 converted to  z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
5 0
3 years ago
How does deposition help explain the layers that are often found in sedimentary rock?
julia-pushkina [17]
Because it is deposition
6 0
3 years ago
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