How would I solve for x and y?

Explain the difference between t(n) and f(x) that makes your answers to parts b and c different

inn [45]

Answer:

t(n) simply means the exact number of times needed to calculate a data of n size while f(x) means the function of the the variable x.

Step-by-step explanation:

3 0

3 years ago

Calculate the mean and median for the data.<br> 24, 26, 40, 27, 37, 32

Olegator [25]

Answer:

mean is 31

median is 29.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the equation of the line perpendicular to y=3/2x-9 and passes through the point (6,2)

lys-0071 [83]

Answer:

y=-2/3x-2

Step-by-step explanation:

slope= -2/3

y=-2/3x+b

enter the given points

2=4+b

b=-2

y=-2/3x-2

6 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago

Two ballpoint pens are selected at random from a box that contains 3 blue pens, 2 red pens, and 3 green pens. If X is the number

Flauer [41]

Answer:

a) f(x,y) = $\frac{\binom{3}{x}\binom{2}{y}\binom{3}{2-x-y}}{\binom{8}{2}}$ ; x = 0, 1 , 2; y = 0, 1 , 2; 0 ≤ x+y ≥ 2

b) = $\frac{9}{14}$

Step-by-step explanation:

joint probability is a function that characterizes the distribution of a random variable. If X and Y be two random variables then the joint probability will be P(X = x, Y=y)

Given Data,

X = The number of blue Pens

Y = The number of red Pens

a)

possible outcomes(X, Y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), (2,0)

Please refer fig. also

total number ways of selecting any 2 pens = $\binom{8}{2}$ = $\frac{8!}{2! 6!}$ =28