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castortr0y [4]
4 years ago
5

Can you explain why estimation can sometimes be helpful when multiplying decimals?

Mathematics
1 answer:
Molodets [167]4 years ago
8 0
Estimation is a quick way to get an approximate result. If all you need is an idea of what the product is, then estimation may be helpful.
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timis mom filled the gas tank in her SUV and spent $72.00 for 22.5 gallons of unleaded gas. What is the constant of proprotional
Dafna1 [17]

Answer:

The constant of proportionality for this situation is 3.2

Step-by-step explanation:

For the given situation we can say that, the amount spent to fill the gas directly varies with the number of gallons of gas filled.

We can say,

\$72.00=k\times 22.5

where k is the constant of proportionality which is nothing but the units rate of gas in dollars per gallon.

Dividing both sides by 22.5 to find k

\frac{72.00}{22.5}=\frac{k\times 22.5}{22.5}

∴ k=3.2

Unit rate of gas = $3.2/gallon

∴ The constant of proportionality for this situation is 3.2.

8 0
3 years ago
How many times larger is the value of the digit 3 in 832,719 than the digit 3 in 824,319?
Leokris [45]

Answer:

100 times larger

Step-by-step explanation:

The digit 3 in 832,719 has the value of 30,000.

The digit 3 in 824,319 has the value of 300.

30,000/300 = 100

3 0
4 years ago
Let X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes. If the random vari
Gelneren [198K]

Answer:

Probability that the wait time is greater than 37 minutes is 0.3474.

Step-by-step explanation:

We are given that the random variable X is known to be exponentially distributed and X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes.

<u><em>Let X = waiting time for a car to pass by on a country road</em></u>

The probability distribution function of exponential distribution is given by;

f(x) = \lambda e^{-\lambda x}  , x >0     where, \lambda = parameter of distribution.

Now, the mean of exponential distribution is = \frac{1}{\lambda}  which is given to us as 35 minutes that means  \lambda = \frac{1}{35}  .

So, X ~ Exp( \lambda = \frac{1}{35} )

Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;

F(x) = P(X \leq x) = 1 - e^{-\lambda x}  , x > 0

Now, Probability that the wait time is greater than 37 minutes is given by = P(X > 37 min) = 1 - P(X \leq 37 min)

  P(X \leq 37 min) = 1 - e^{-\frac{1}{35} \times 37}        {Using CDF}

                         = 1 - 0.3474 = 0.6525

So, P(X > 37 min) = 1 - 0.6525 = 0.3474

Therefore, probability that the wait time is greater than 37 minutes is 0.3474.

4 0
3 years ago
(3/12) + 1/12 - 11/12
worty [1.4K]

Answer:

-7/12

Step-by-step explanation:

1. so you add (3/12) + 1/12= 4/12

2. 4/12-11/12=  -7/12

Plz name me the brainliest

8 0
3 years ago
Need to know these two problems
Arturiano [62]
Multiple 16 by 1 and then divide by 8 which makes 2.

Multiply 36 by 1 and then divide by 9 which makes 4.
5 0
3 years ago
Read 2 more answers
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