What is the mean absolute deviation of the following set of data?<br> 4<br> 6<br> 6<br> 2.

Steve and Ben have some money in their wallets. If Steve gives Ben $3, Ben will have twice as much as Steve. If Ben gives Steve

Zigmanuir [339]

Answer:

Money owned by Steve = $11

Money owned by Ben = $13

Step-by-step explanation:

Let x denotes the money owned by Steve and y denotes the money owned by Ben.

Then, Steve gives $3 to Ben

Money left with Steve = x-3 and Ben = y+3

Now, Ben will have twice as much as Steve.

⇒ y+3 = 2(x-3) .............(1)

If Ben gives Steve $7

Money left with Steve = x+7 and Ben = y-7

Then, the amount Ben has will be one-third that of Steve’s.

⇒ y-7 = $\frac{1}{3}$ (x+7) ..........(2)

Solving equation (1) and (2) by elimination method, we get

x = 11 and y = 13

⇒ Money owned by Steve = $11

Money owned by Ben = $13

5 0

3 years ago

Find a polynomial f(x) of degree 3 that has the following zeros.

JulijaS [17]

Answer:

f(x) = x(x - 4)(x + 3)

Step-by-step explanation:

given f(x) with zeros x = a and x = b then the corresponding factors are

(x - a) and (x - b)

f(x) is then the product of these factors

Given zeros are x = 4, x = 0, x = - 3 then the factors are

(x - 4) , (x - 0), (x - (- 3)) , that is

(x - 4) , x , (x + 3)

f(x) = x(x - 4)(x + 3)

3 0

2 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co

Ira Lisetskai [31]

Answer:

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$ .

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0

3 years ago

Hellllllp see the pictures

kupik [55]

A. All remaining numbers are in Q because it goes "even" left, and "odd" right.

B. Probability of it not being in Q is 1/2

3 0

3 years ago

Which could be the original expression?

miss Akunina [59]

In order to answer this I have to see a visual

5 0

3 years ago

What is the mean absolute deviation of the following set of data? 4 6 6 2.