Ask question

Ask question

All categories

Tcecarenko [31]

3 years ago

11

SOMEONE PLEASE HELP ME ASAP PLEASE !!!!!

1 answer:

Masja [62]3 years ago

3 0

Answer:21

Step-by-step explanation:

Find the median. Separate everything above the median into two groups. Take the group with the larger numbers we’ll call this the upper quartile. Find the median of the larger group or upper quartile. Take the group with the smaller numbers, which we’ll call the lower quartile. Find the median of the smaller group or lower quartile. Subtract the median of the lower quartile from the median of the upper quartile and you will find the inner quartile range. In this case, it is 21

You might be interested in

What is 15% of 12 meters

melomori [17]

Think of the "of" as a multiplication sign:
(.15)(12)= 1.8 meters
There you go! hope this helps with this problem and future ones.

7 0

3 years ago

Read 2 more answers

Ruth wants to purchase a new video game that costs $59.00 and several bags of chips that sell for $1.99 each. She can spend no m

allsm [11]

Answer:

1.99c + 59 ≥ 75, where c ≥ 8.04

Step-by-step explanation:

8 0

2 years ago

For a ride on a rental scooter, Deon paid an $8 fee to start the scooter plus 6 cents per minute of the ride. The total bill for

faltersainse [42]

Answer:

66 minutes

Step-by-step explanation:

11.96 - 8 =3.96

Every 10 minutes Deon pays 60 cents because 6 times 10 = 60 and every 6 minutes Deon is paying 36 cents

10 min = 60 cents

20 min = $1.20

40 min = $2.40

50 min = $3.00

60 min = $3.60

6 min = 36 cents

Then you do 3.60 + 36 which equals $3.96

How to figure how the minutes is just 60 + 6 = 66 minutes

7 0

2 years ago

- 5+x=-8 how do solve this equation

zloy xaker [14]

Add 5 to both sides to get x alone and you will have your answer which is x= -3

5 0

3 years ago

Read 2 more answers

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago