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rjkz [21]
3 years ago
13

You have a bag of marbles that contains, 7 red, 4 green, and 5 yellow

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

i hope my answer help you

Step-by-step explanation:

the probability to pull out a green marble=4/16 ( simplest form =1/4)

the probability to pull out a green marble after replace the previous marble=3/15 (simplest form =1/5)

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The ratio 6:7 can also be written as 7/6 <br> True or false
zloy xaker [14]
I think the answer is false✨
5 0
3 years ago
Please answer
Vladimir [108]

Answer:

Step-by-step explanation:

2kg/m² convert to g/100cm²

1 kg=1000 g

2kg=  ???

2*1000=2000kg

m      ⇒      100cm

m²     ⇒      10000 cm²

2kg/m²=2*1000/(10000/100) cm²

=2000/100 cm²

<h2>4- 250kg/hectare</h2>

250*1000=250000

1 hectare=10000 m²

250000/10000 g/m²

3 0
3 years ago
What are the factors for 24, 64 and 88
DiKsa [7]
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 64: 1, 2, 4, 8, 16, 32, 64<span>

Factors of 88: </span>1, 2, 4, 8, 11, 22, 44, 88
<span>
The common factors of 24, 64, and 88 are: 1, 2, 4, and 8.</span>
8 0
3 years ago
Read 2 more answers
80% of what number is 94
lapo4ka [179]
What i would do is make it an equation
80% would turn to .80
.80 * (of) X = (is) 94
.80 * X = 94

next you would /.8 from both sides making the end result- 

X=117.5 

then plug it in to check
.80 * 117.5 = ?
.80 * 117.5 =94 

hope this helped! 
5 0
3 years ago
Read 2 more answers
Jill bought a car for $50,000. Her car depreciates at a rate of 10% a year. create an equation for the problem
Vadim26 [7]

Answer:

The value of car after n years at the depreciation rate is                              $ 50,000 (0.9)^{n}  .

Step-by-step explanation:

Given as :

The cost of the car that Jill bought = $ 50,000

The depreciation rate of car value = r = 10 % a years

Let The car after n years of depreciation = $ A

Now, According to question

The cost of car after n years of depreciation = initial cost of car × (1 - \dfrac{\textrm rate}{100})^{\textrm time}

Or, $ A = $50,000 × (1 - \dfrac{\textrm r}{100})^{\textrm n}

Or, $ A = $50,000 × (1 - \dfrac{\textrm 10}{100})^{\textrm n}

Or, $ A = $50,000 × (\frac{90}{100})^{n}

I.e $ A = $50,000 × (0.9)^{n}

So, value of car after n years = $ 50,000 (0.9)^{n}

Hence The value of car after n years at the depreciation rate is                      $ 50,000 (0.9)^{n}  . Answer

6 0
3 years ago
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