Write the first 10 terms of a sequence who’s first term is -10 and who is common difference is -2

Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997, However in march 1997 the price of the tea increased by 25% and

Irina18 [472]

Answer: tea = 15 rupees per kg

sugar= 3 rupees per kg

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":

2 t + 3 s =39 (a)

Where:

t= price of 1 kg of tea
s = price of 1 kg of sugar

"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30. "

2(t1.25)+3(s1.2) = 48.30 (b)

Solving for t in (b)

2t =39-3s

t = (39 -3s)/2

t = 19.5-1.5s

Replacing the value of t in (b)

2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30

2x ( 24.375 -1.875s) +3.6s =48.30

48.75 -3.75s+3.6s= 48.30

48.75-48.30 = 3.75s-3.6s

0.45= 0.15s

0.45/0.15 =s

3 =s

Replacing the value of s in (a)

2 t + 3 (3) =39

2 t + 9 =39

2 t =39 -9

2 t =30

t = 30/2

t= 15

Prices in january:

tea = 15 rupees per kg

sugar= 3 rupees per kg

Feel free to ask for more if needed or if you did not understand something.

4 0

3 years ago

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A closet contains 8 pairs of pants, 10 shirts, and 4 pairs of shoes. How many leaves on a tree diagram would represent all possi

EleoNora [17]

The answer is, there are 320 possible combinations of outfits

7 0

3 years ago

Lake Michigan's volume is approximately 1,180 cubic miles and its surface area is approximately 14,332,090 acres. The 2015 water

galben [10]

Answer:

a.) depth of the lake = (volume of the lake) / area of lake

= 173,693,585,360,000 / 624,305,840,400 = 278.22 feet

b.) 11 inches = 0.9167 feet

Water gained in cubic feet gained by the lake

= 624,305,840,400 $\times$ 0.9167

= 572,282,434,719.5 cubic feet

c.) the percentage change in the lake volume over that year

= 572,282,434,719.5 cubic feet / 173,693,585,360,000 cubic feet

= 0.0033

= 0.33%

Step-by-step explanation:

i) acres to square feet : 1 acre = 43560 square feet

therefore 14,332,090 acres = 624,305,840,400 square feet

ii) 1 mile = 5280 feet

1 cubic mile = 5280 $\times$ 5280 $\times$ 5280 = 147,197,952,000 cubic feet

therefore 1180 cubic miles = 173,693,585,360,000 cubic feet

a.) depth of the lake = (volume of the lake) / area of lake

= 173,693,585,360,000 / 624,305,840,400 = 278.22 feet

b.) 11 inches = 0.9167 feet

Water gained in cubic feet gained by the lake

= 624,305,840,400 $\times$ 0.9167 = 572,282,434,719.5 cubic feet

c.) the percentage change in the lake volume over that year

= 572,282,434,719.5 cubic feet / 173,693,585,360,000 cubic feet

= 0.0033

= 0.33%

7 0

4 years ago

From point A(3,-2), walk 4 spaces in the negative x direction, and 3 spaces in the positive y direction. Use a paper cutout to l

Zolol [24]

Answer: 3,-2Step-by-step explanation: ... Use a paper cutout to label the point you reach as. B. What are the coordinates of point B?

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

3 years ago