How can I solve this problems?

Mathematics

1 answer:

nadezda [96]2 years ago

7 0

1)
Julia = Sj = 30,500 + 500n
Aysha = Sa = 26,000 + 1000n

when will
Sj = Sa
30,500 + 500n = 26,000 + 1000n
30,500 - 26,000 = 1000n - 500n
4,500 = 1000n - 500n
4,500 = 500n
n = 9
so they will have same salary after 9 years

2) X + Y = 8000
5%X + 3.25%Y = 312.5

X = 8000 - Y
0.05 ( 8000 - Y ) + 0.0325Y = 312.5
400 - 0.05Y + 0.0325Y = 312.5
400 - 0.0275Y = 312.5
400 - 312.5 = 0.0275Y
87.5 = 0.0275Y
Y = 3,181.81
X = 8000 - Y = 8000 - 3,181.81 = 4,818.19

3) X + Y = 3050
8%X + 7.5%Y = 234

X = 3050 - Y
0.08(3050 - Y) + 0.075Y = 234
244 - 0.08Y + 0.075Y = 234
244 - 234 = 0.025Y
Y = 400
X = 3050 - Y = 3050 - 400 = 2650

You might be interested in

A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If

Blababa [14]

Answer: Our required probability is $\dfrac{1}{2}$

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = $\dfrac{1}{3}$

So, Probability of getting head from fair coin = $\dfrac{1}{2}$

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

$\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}$