Answer:
Their are caculators online search the type of math it is then put caculator for example exponents caculator
Step-by-step explanation:
Answer:
Siti's money = RM 430
David = RM 1,290
Farid = RM 280
Step-by-step explanation:
Let
Siti's money = x
David = 3x
Farid = x - 150
Total of their money = RM 2 000
x + 3x + (x - 150) = 2000
4x + x - 150 = 2000
5x = 2000 + 150
5x = 2,150
x = 2,150/5
x = RM 430
Siti's money = x
= RM 430
David = 3x
= 3(430)
= RM 1,290
Farid = x - 150
= 430 - 150
= RM 280
Answer:
Null hypothesis:
Alternative hypothesis:
The statistic to check the hypothesis is given by:
And is distributed with n-2 degrees of freedom
And the statistic to check the significance of a coeffcient in a regression is given by:

For this case is importantto remember that t1 and p value for test of slope coefficient is the same test statistic and p value for the correlation test so then the answer would be:
Always
Step-by-step explanation:
In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:
Null hypothesis:
Alternative hypothesis:
The statistic to check the hypothesis is given by:
And is distributed with n-2 degrees of freedom
And the statistic to check the significance of a coeffcient in a regression is given by:

For this case is importantto remember that t1 and p value for test of slope coefficient is the same test statistic and p value for the correlation test so then the answer would be:
Always
We know that the area is 20.
Since 20 is a small number: lets list out possible combinations of lengths and widths.
1 * 20
2 * 10
4 * 5
L = 7 + 3w
lets see which on makes sense.
L = 7 + 3w
20 = w7 + 3w^2
3w^2 + 7w -20 = 0
(3w - 10)(w - 2)
w can equal 10/3 or 2.
So the dimensions: are Width = 2 Length = 10
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
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