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3 years ago

3g + 4g How do you this? Is there a simple way??

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

5 0

They're like terms. So it will be 7g as you plus the co-efficient together and you 7 and the g you add it to it.

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Lucas makes a flag it has 6 equal parts five of the parts are red what fraction of the flag is red

deff fn [24]

One part of six parts is red, so just make it a fraction like so: 1/6 because 1 side is red out of six total parts.

8 0

3 years ago

(1.5×10^3)+(2.5×10^2) in standard form

Tcecarenko [31]

Answer: 2.65·10^-2

Step-by-step explanation. : 1.5·10^-3= 0.0015 and 2.5·10^-2= 0.025.

Sum is 0.0265

7 0

3 years ago

detrermine weather the table represents an exponential growth function, and exponential decay function, or neither... PLEASE HEL

Helen [10]

Answer:

The y values are decreasing, I would say it's decay.

7 0

3 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

What is the percentage increase if your salary from$85 to $100 per week round to the nearst percent

natali 33 [55]

To find what percent of something something is you first divide the part by the whole (85/100=0.85) and then multiply that by 100 (0.85 * 100=85) this means you get 85%

6 0

3 years ago