Let the numbers be n, n+2, n+4
Sum equals too= 13+2(n+4), which is 2n+21
a) Equation--> n+n+2+n+4= 2n+21
b) Solution--> 3n+6= 2n+21
=> n= 15
c) Second number--> 17 (15+2)
Third number--> 19 (15+4)
d) 15+15+2+15+4=30+21
=> 51= 51
So, the equation is true.
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Answer:
Seven million two thousand five hundred in standard form is 7002500
Tfgycg hcjg hg bgguvv hcjugb bybvh
Answer:
The answer is "Type 1 error".
Step-by-step explanation:
The error of type I, frequently known as a 'false positive': its error in judgment that perhaps a null hypothesis is simply rejected. This is the mistake of accepting a possible (actual interest hypothesis) hypothesis whenever the results could be attributed to chance since the researchers deny, if valid, the null hypothesis.
Answer:
i. terms = 5xy(z)^2 , -3zy
coefficient = 5 and -3
ii. terms = 1 , X ,x^2
coefficient = 1 , 1
iv. terms = 3 , -pq , qr, -rp
coefficient = -1 ,1 , -1
v. terms = X/2 , y/2 , -xy
coefficient= 1/2 ,1/2 and -1
