I need help to solve this problems can somebody help me
Compare the numbers using <, >, or =. 0.78 ___ 0.708 < > =
In triangle ABC,
AC = 12/ (sin30) = 12 / (1/2) = 24
DC = 24-x
DB = DC tan 30 = (24-x) tan30 <span>=(24−x)/</span><span>√3
</span>
In triangle ADB using Pythagorean Theorem<span><span>x2</span>+((24−x)/<span>√3</span><span>)2</span>=<span>12^2</span></span><span><span>x2</span>+(24−x<span>)^2</span>/3=<span>12^2</span></span><span>3<span>x2</span>+(24−x<span>)^2</span>=432</span><span>4<span>x2</span>−48x+576=432</span><span>4<span>x2</span>−48x+144=0</span><span><span><span>x2</span>−12x+36=0
x1 = x2 =6
AD = AC - DC = 24- (24-x) = 6</span></span>
Complete Question
The complete question is shown on the first uploaded image
Answer:
the null hypothesis is 
the alternative hypothesis is 
The test statistics is 
The p-value is 
so
Step-by-step explanation:
From the question we are told that
The population mean is 
The sample size is n= 38
The sample mean is 
The standard deviation is 
Generally the null hypothesis is
the alternative hypothesis is
Generally the test statistics is mathematically evaluated as
substituting values
The p-value is mathematically represented as
From the z- table
So
Answer:
in this method a variable is expressed in terms of another variable from one equation and it is substituted in remaining equation.
Step-by-step explanation:
solve:x+2y=9 and 3x-y=13.
x+ 2y=9…be the first equation
3x-y=13…be the second equation
from first equation
x=9-2y…be the third equation substituting value of x from third equation in second equation, we get
3(9-2y)-y=13
or, 27-6y-y=13
or, y=2
now,
substituting value of y in third equation, we get,
x=9-2×2
=9-4
=5
the required value of x and y are 5&2
