Answer:
<em>Hence the daughter's present age is 15 years</em>
<em>The fathers present age is 35 years</em>
<em></em>
Step-by-step explanation:
Let the present age of daughter be x
Let the present age of father be y
5 years ago;
Daughter's age = x - 5
Fathers age = y - 5
If the present age of father is thrice as old as the age of daughter 5 years ago, then;
y - 5 = 3(x-5)
y - 5 = 3x-15
y = 3x - 10 .... 1
In 5 years time;
Daughters age = x + 5
Fathers age = y + 5
If the age of father will be twice the age of his daughter in 5 years time then;
y+5 = 2(x+5)
y+5 = 2x + 10
y = 2x + 5 .....2
Equate 1 and 2;
3x - 10 = 2x + 5
3x - 2x = 5 + 10
x = 15
Since y = 2x + 5
y = 2(15) + 5
y = 35
<em>Hence the daughter's present age is 15 years</em>
<em>The fathers present age is 35 years</em>
<em></em>
1ft=12in
1ft/12in=1
3.5ft=?in
1ft:12in=3.5ft:?in
1ft/12in=3.5ft/?in
times both sides by 12?in
1?=42in
answer is 42 in
or you could just do
3.5 times 12=42 inches
Answer:
55 degrees is more unusual in July as it is away from mean temperature as compared to January.
Step-by-step explanation:
We are given the following information in the question:
January:
Mean, μ = 36 degrees
Standard Deviation, σ = 10 degrees
July:
Mean, μ = 74 degrees
Standard Deviation, σ = 8 degrees
We calculate z scores corresponding to x = 55.
Formula:

January:

July:

Thus, 55 degrees is more unusual in July as it is away from mean temperature as compared to January.
Answer:
f(n) = 1
Step-by-step explanation:
Another answer for -3. FxN= -3