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luda_lava [24]
2 years ago
4

Fido weighs 26 pounds more than fifi. fifi weighs 12 pounds less than rover. if the sum of their weights is 71, how much does fi

fi weigh?
Mathematics
2 answers:
Ronch [10]2 years ago
8 0
Hey there,
Let's say Rover's weight is x,
Rover = x
Fifi = x - 12 
Fido = x - 12 + 26
x + x - 12 + x - 12 + 26 = 71
3x - 12 - 12 + 26 = 71
3x + 2 = 71
3x = 71 - 2
3x = 69
x = 69 / 3
  = 23
Fifi = 23 - 12
      = 11 pounds

Hope this helps :)) It would be great if you would mark my answer as the brainliest ;)

<em>~Top♥</em>
PIT_PIT [208]2 years ago
5 0
Here you go:

Let's say fido is A, fifi is B, and Rover is C.
#1. a=26+b
#2.c=12+b
a+b+c=71
Replace a and c with the equations #1 and #2 to get the equation 26+b+12+b+b=71. Simplify this to get 3b+38=71, and then you subtract 71 by 38 to eliminate 38, and you are left with 3b=33. Divide the 3 to 33 to get your final answer, where Fifi is 11 pounds. Hope this helps!

~Brainliest
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An alloy is composed of nickel, zinc, and copper in a 7:2:9 ratio. How many kilograms of each metal is needed to make 3.8 kg of
Tamiku [17]
Ni : Zn: Cu = 7:2:9 =7x : 2x : 9x

7x+2x+9x = 3.8
18x=3.8
x=3.8/18=1.9/9≈0.21

Ni:   7x=7*0.21=1.47 ≈1.5 kg
Zn:  2x=2*0.21=0.42≈0.4 kg
Cu: 9x=9*0.21=1.89 ≈1.9 kg

Check 1.5+0.4+1.9 =3.8 True
6 0
2 years ago
Study the steps used to solve the equation. Given: StartFraction c Over 2 EndFraction minus 5 equals 7 Step 1: StartFraction c O
Bas_tet [7]

Answer:

  1. addition property of equality
  2. integers are closed to addition
  3. identity element
  4. multiplication property of equality
  5. commutative property of multiplication; reals are closed to multiplication; identity element

Step-by-step explanation:

<u>Given</u>:

  c/2 -5 = 7

  Step 1: c/2 -5 +5 = 7 +5

  Step 2: c/2 +0 = 12

  Step 3: c/2 = 12

  Step 4: 2(c/2) = 12(2)

  Step 5: c = 24

<u>Find</u>:

  The property that justifies each step of the solution.

<u>Solution</u>:

  Step 1: addition property of equality (lets you add the same to both sides)

  Step 2: integers are closed to addition

  Step 3: identity property of addition (adding 0 changes nothing)

  Step 4: multiplication property of equality

  Step 5: closure of real numbers to multiplication; identity property of multiplication

_____

It is hard to say what "property" you want to claim when you simplify an arithmetic expression. Above, we have used the property that the sets of integers and real numbers are closed to addition and multiplication. That is, adding or multiplying real numbers gives a real number.

In Step 5, we can rearrange 2(c/2) to c(2/2) using the commutative property of multiplication. 2/2=1, and c×1 = c. The latter is due to the identity element for multiplication: multiplying by 1 changes nothing.

Apart from the arithmetic, the other properties used are properties of equality.  Those let you perform any operation on an equation, as long as you do it to both sides of the equation. The operations we have performed in this fashion are adding 5 and multiplying by 2.

8 0
2 years ago
Read 2 more answers
I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?
san4es73 [151]
(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
8 0
2 years ago
Jane invested £4000 for 3 years with an interest rate of 1.5%. What was her investment worth at the end of this period?
ddd [48]

Answer:

$4182.7

Step-by-step explanation:= 4000*(1.5%)*3

Year 1= 4000*(100%+1.5%)= 4060

Year 2= 4060*(100%+1.5%)= 4120.9

Year 3= 4120.9*(100%+1.5%)= 4182.7

6 0
2 years ago
Read 2 more answers
A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol
bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that p = 0.25

25 incoming calls.

This means that n = 25

a. What is the probability that at most 4 of the calls involve a fax message?

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4). Then

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

P(X \leq 4) + P(X > 4) = 1

From a), P(X \leq 4) = 0.214)

Then

P(X > 4) = 1 - 0.214 = 0.786

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0
2 years ago
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