Helppppppp pls this is due in a few hours

5pts and the person with the correct answer gets brainliest

WITCHER [35]

Answer:

(a) 5

(b) 3rd quartile

(c) 2nd quartile

Step-by-step explanation:

There are four quartiles in a box plot (or box and whisker plot).

The 1st is from 5 to 13

The 2nd is from 13 to 35

The 3rd is from 35 to 40

The 4th is from 40 to 48

(a) The lowest that this box plot goes is 5 (between the 4 and 6)

(b) 3rd quartile — we can see that the left and right borders are close together, meaning that the data values for this quartile are close together, and therefore concentrated

(c) 2nd quartile — we can see that this quartile is the widest, and therefore the data values are spread out

8 0

2 years ago

In order to circumscribe a circle about a triangle, the circle's center must be placed at the incenter of the triangle.

zimovet [89]

The answer would be False!

7 0

3 years ago

A pink paint mixty uses 4 cups of white paint for every 3 cups of red paint

s2008m [1.1K]

Answer:

112

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago

Complete the table for the following equation then graph the line. (Show steps in the middle column of table) y=-3x-2

kotegsom [21]

Answer:

Step-by-step explanation:

3 0

2 years ago