Answer:
1 cm : 2m
1 cm = 2m
118m
Step-by-step explanation:
A scale drawing is a reduced form in terms of dimensions of an original image / building / object
the scale drawing is usually reduced at a constant dimension
scale of the drawing = original dimensions / dimensions of the scale drawing
18 / 9 = 2 : 1
59 x 2 = 118
Answer:
Test statistic z= 1.708
Step-by-step explanation:
given that a television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot.
Sample size n = 827
Sample proportion p = 
(right tailed test)
This is because they would be interested if more than half viewers are ready for the change
Test statistic = Z = p difference/std error
p difference = 0.0297
Std error =
Z =1.708
p valu e= 0.9562
Since p >0.05 we fail to reject H0
Test statistic z= 1.708
Answer:
a) 1+2+3+4+...+396+397+398+399=79800
b) 1+2+3+4+...+546+547+548+549=150975
c) 2+4+6+8+...+72+74+76+78=1560
Step-by-step explanation:
We know that a summation formula for the first n natural numbers:
1+2+3+...+(n-2)+(n-1)+n=\frac{n(n+1)}{2}
We use the formula, we get
a) 1+2+3+4+...+396+397+398+399=\frac{399·(399+1)}{2}=\frac{399· 400}{2}=399· 200=79800
b) 1+2+3+4+...+546+547+548+549=\frac{549·(549+1)}{2}=\frac{549· 550}{2}=549· 275=150975
c)2+4+6+8+...+72+74+76+78=S / ( :2)
1+2+3+4+...+36+37+38+39=S/2
\frac{39·(39+1)}{2}=S/2
\frac{39·40}{2}=S/2
39·40=S
1560=S
Therefore, we get
2+4+6+8+...+72+74+76+78=1560
Do you have a picture to go with this question ?
Answer:
See below
Step-by-step explanation:
<u>An invalid solution.</u>
When you use the quadratic formula, for example, to solve for 't', time, you will often find one of the values of 't' to be NEGATIVE value for time.....this would be an <u>extraneous</u> solution.
Sometimes when working with LOG equations you may fin an answer which is negative.....you cannot have a LOG of a negative number...THAT solution would be EXTRANEOUS
ETC
SO, always check your answers to see if they 'work' for your problem.
