If g (x)=2x + 1, find g (a + h)- g (a).

Blake and three friends meet for lunch. His friends all get the same thing, but Blake gets a different lunch that cost $6. Write

sladkih [1.3K]

Answer:

Step-by-step explanation:

Friends lunch=8+8+8=$32

Blake's lunch=$6

32+6=$40 total

6 0

3 years ago

10. 0.75,1/2, 0.4,and 1/5 from least to greatest

Pavlova-9 [17]

Let's turn these into decimals.

so, 10.

0.75

0.50=1/2

0.40

0.20=1/5

In order from least to greatest, these would be:

0.2, 0.4, 0.5, 0.75, 10

5 0

3 years ago

How to write 8/15 as a decimal?

Murrr4er [49]

Answer:

0.53

Step-by-step explanation:

Basically just go to your calculator and put in 8 divded by 15

3 0

3 years ago

Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and

ryzh [129]

Answer:

a) X1+X2 have distribution Bi(n1+n2, p)

b)

P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

4 0

3 years ago

true or false If x represents a random variable with mean 114 and standard deviation 40, then the standard deviation of the samp

Goshia [24]

Answer:

$\mu = 114, \sigma = 40$

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sampling distribution would be:

$\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4$

So then the answer is TRUE

Step-by-step explanation:

Let X the random variable of interest and we know that the true mean and deviation for this case are given by:

$\mu = 114, \sigma = 40$

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard deviation for the sampling distribution would be:

$\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4$

So then the answer is TRUE

6 0

3 years ago