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3 years ago

Solving systems of equations by graphing

Mathematics

1 answer:

abruzzese [7]3 years ago

4 0

X= 2.8
Y= 4.4
( 2.8 , 4.4 )

It’s been a long time since I’ve done this so this is the best I could do

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Mr. Gurung sold a bicycle for Rs 18,816 and made a profit of 12%. if he wants

LenaWriter [7]

Answer:

Rs. 23520

Step-by-step explanation:

Given that :

At 12% profit ; sales price = 18,816

At (12 + 3)% profit ; sales price = x

Hence,

0.12 = 18,816

0.15 = x

Cross multiply ;

0.12x = 0.15 * 18816

0.12x = 2822.4

x = 2822.4 / 0.12

x = 23520

Hence, sales price for a 3% increaa in profit should be Rs. 23520

5 0

2 years ago

if a rectangle has an area of 30 square meters and a perimeter of 34 meters . what is the dimensions of the rectangle?

timama [110]

P=2(L+W) and P=34 so

2(L+W)=34

L+W=17 so we can say

L=17-W

A=LW using L from above

A=(17-W)W

A=17W-W^2 and A=30 so

30=17W-W^2

W^2-17W+30=0

W^2-2W-15W+30=0

W(W-2)-15(W-2)=0

(W-15)(W-2)=0

So the dimensions of the rectangle are 15 meters by 2 meters.

7 0

3 years ago

Question 7 (1 point)<br> Solve: 8x = 56

Hunter-Best [27]

Answer:

x = 7

Step-by-step explanation:

8x = 56

To solve this, we must simplify. To do this, we must divide each side by 8. This, in term, will give us the value of x.

8x = 56

---- ----

8 8

56/8 = 7

x = 7

Our answer is x = 7

5 0

3 years ago

I will give Brainly ! help please its geometry I will give brainly <br> Thank you for your time.

Cerrena [4.2K]

Answer:

I guess reflexive property........

4 0

1 year ago

Read 2 more answers

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

2 years ago