Find the equation of a line that contains the points (−3,−1) and (−4,−7). Write the equation in slope-intercept form

Help me.... IF A=1, B=2, AND C=929, WHATS C/A x B + B x C

morpeh [17]

I got 1,727,940
Plug in the numbers

7 0

3 years ago

Read 2 more answers

Correct answer gets brainliest

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

<em>I had the same question and the answer was B.</em>

7 0

3 years ago

The proportion of brown M&M's in a milk chocolate packet is approximately 14%. Suppose a package of M&M's typically cont

marshall27 [118]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the scenario above :

A) State the random variable.

The random variable is the p opoertion of brown M&M's in a milk chocolate packet.

B.) Argue that this is a binomial experiment.

Each trial is independent for a total number of 52 trials with a set probability of success at 0.14

C) probability that 6 M&M's are brown:.

P(x) = nCx * p^x * (1-p)^(n-x)

p = 0.14 ; (1 - p) = 0.86 ; n = 52 ; x = 6

P(x = 6) = 52C6 × 0.14^6 × 0.86^46

= 20358520 × 0.00000752954 × 0.00097035078

= 0.1487

D) P(x =25)

P(x = 25) = 52C25 × 0.14^25 × 0.86^27

= 477551179875952 × 449.987958058*10^(-24) × 0.01703955245

= 0.00000000366

E) P(x = 52)

P(x = 52) = 52C52 × 0.14^52 × 0.86^0

= 1 × 3968.78758299*10^(-48) × 1

= 3968.78758299*10^(-48)

F) yes it would be unusual, because such probability is extremely low. However, if a huge or substantial number of trials such may occur

3 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$