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melomori [17]
3 years ago
15

Subtract : 8,878-2,314 5.4 from 12 ( the difference is.....

Mathematics
1 answer:
mixas84 [53]3 years ago
7 0

Answer:

1st question : 6564

2nd question : 6.6

Step-by-step explanation:

We have to subtract 8,878-2,314.

We will subtract the smaller number from greater number.

So, the answer will be = 6564

We also have to subtract 5.4 from 12 .

Hence, the difference will be = 6.6

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I<br>buy<br>book for $8,i<br>sell it for<br>$ 3.25.<br>How much<br>of<br>loss<br>do<br>i<br>make ?​
Anuta_ua [19.1K]

Answer:

Loss = $ 4.75

Step-by-step explanation:

Loss = cost price -selling price

You bought a book will be cost price and when u r selling it it will be selling price

loss = 8-3.25

      = 4.75

Loss = $ 4.75

4 0
3 years ago
Read 2 more answers
Consider the quadratic equation x3 = 48-5. How many solutions does the equation have?
Arturiano [62]

Answer:

 you would  have 84-.10 I think

Step-by-step explanation:

5 0
3 years ago
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Mrs Perry shares out 15 biscuits between Gemma and Zak in the ratio 1:2
Bad White [126]
<span>1x + 2x = 15 3x = 15 3x/3 = 15/3 x = 5 Gemma = 1x, Zak = 2x therefore Gemma = 5 and Zak = 10</span>
5 0
3 years ago
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If f(x)=2​x+1, then f(3)= ?
Phantasy [73]

Answer:

9

Step-by-step explanation:

Input 3 for x

f(x)=2^3+1

f(3)=2^3+1

f(3)=8+1

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4 0
3 years ago
Suppose the lifetime, in years, of a motherboard is modeled by a Gamma distribution with parameters α=80 and λ=4. Use the Centra
timurjin [86]

Answer:

Therefore there's a 99.99% probability the motherboard of your new computer will last for at least 15 years.

Step-by-step explanation:

This is the general idea to solve the problem.

Suppose that the mean and variance of the your distribution are .\mu , \sigma respectively. Then, according to the problem you are looking for the probability.

P(X \geq 15) = 1 - P(X

Consider then the following random variable.

T  =   X_1 + X_2 + .... + X_{14}

Using the central limit theorem  T distribution will be close to normal, and its mean and variance will be  14\mu , 14\sigma  , respectively. Therefore you just have to find the probability that a normally distributed  random variable with that mean and that variance which I just mentioned is less than 14.

For this case we have that

\mu = \alpha / \gamma  = 20 \\\sigma = \alpha(\alpha+1) / \gamma^2  -  (\alpha/ \gamma)^2  = 5

Then you have that

14\mu = 280\\14\sigma = 70\\

and we have that if  N is a normally distributed random variables with mean 280 and variance 70 we have that

P(N \leq 14 ) = 0.0001

the actual probability we are looking for is

1-P(N\leq14) =  1-(0.0001) = 0.9999

Therefore there's a 99.99% probability the motherboard of your new computer will last for at least 15 years.

5 0
3 years ago
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