Answer:
Daughter age = 3 years
Step-by-step explanation:
Let x be the age of the women and y be the age of the daughter.
Given:
After five year the sum of the women and daughter age = 40
![(x+5)+(y+5)=40](https://tex.z-dn.net/?f=%28x%2B5%29%2B%28y%2B5%29%3D40)
At present the sum of the women and daughter age
![x+5+y+5=40](https://tex.z-dn.net/?f=x%2B5%2By%2B5%3D40)
![x+y+10=40](https://tex.z-dn.net/?f=x%2By%2B10%3D40)
![x+y=40-10](https://tex.z-dn.net/?f=x%2By%3D40-10)
--------------(1)
So the sum of the present age is ![x+y=30](https://tex.z-dn.net/?f=x%2By%3D30)
The difference in their present age is 24 years.
![x-y=24](https://tex.z-dn.net/?f=x-y%3D24)
![x=24+y](https://tex.z-dn.net/?f=x%3D24%2By)
Now we substitute x value in equation 1.
![(24+y)+y=30](https://tex.z-dn.net/?f=%2824%2By%29%2By%3D30)
![24+2y=30](https://tex.z-dn.net/?f=24%2B2y%3D30)
![2y=30-24](https://tex.z-dn.net/?f=2y%3D30-24)
![2y=6](https://tex.z-dn.net/?f=2y%3D6)
![y=\frac{6}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B6%7D%7B2%7D)
![y=3\ years](https://tex.z-dn.net/?f=y%3D3%5C%20years)
Therefore, the daughter age is 3 years.
Answer:
A
Step-by-step explanation:
the height a creates with half of the baseline (5) and a leg (10) a right-angled triangle, and we can use Pythagoras to calculate a.
c² = a² + b²
c being the Hypotenuse (the side opposite of the 90° angle, so in our case the 10 side).
10² = a² + 5²
100 = a² + 25
75 = a²
a = sqrt(75) = sqrt(3×25) = 5×sqrt(3)
Answer:
1π
Step-by-step explanation:
suppose the radius of semicircle P is r,
then the radius of semicircle Q = (r+d)/2 ... d≤r
radius of semicircle R = (r-d)/2
area P = 1/2 (r)²π
area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π
area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π
shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π
= ((r² - d²)/4) * π
because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0
therefore the shaded area = ((2² -0²)/4)*π = 1π
<u>Answer:</u>
The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744
<u>Solution:</u>
Total number of coils = number of good coils + defective coils = 88 + 12 = 100
p(getting two good coils for two selection) = p( getting 2 good coils for first selection )
p(getting 2 good coils for second selection)
p(first selection) = p(second selection) = ![\frac{\text { number of good coils }}{\text { total number of coils }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7B%20number%20of%20good%20coils%20%7D%7D%7B%5Ctext%20%7B%20total%20number%20of%20coils%20%7D%7D)
Hence, p(getting 2 good coil for two selection) = ![\frac{88}{100} \times \frac{88}{100} =\bold{0.7744}](https://tex.z-dn.net/?f=%5Cfrac%7B88%7D%7B100%7D%20%5Ctimes%20%5Cfrac%7B88%7D%7B100%7D%20%3D%5Cbold%7B0.7744%7D)
Answer:
1 : 30
Step-by-step explanation:
300mm:9m
We need to change the meters to mm
1 meter is 1000 mm
so 9 m is 9000 mm
300mm:9000mm
Divide both sides by 300
300mm/300 : 9000/300
1 : 30