Ask question

Ask question

All categories

2 years ago

7

It has been reported that 48% of teenagers play video games on their phones. A random sample of 60 teenages is drawn. Find the p

robability that the proportion of teenagers in the sample who play videos games on their phone is less than 0.489 .

1 answer:

Ivanshal [37]2 years ago

3 0

Answer:

skakaaja

Step-by-step explanation:

aamnanananan

You might be interested in

Which statements are true about the circle shown below? Select two options.

AfilCa [17]

Answer:

C,D

Step-by-step explanation:

E d g e n u i t y

5 0

3 years ago

Read 2 more answers

Which number has an absolute value that is greater then 50?

Lelu [443]

Any number that greater that 50 or any Number that if on the left of -50

7 0

3 years ago

Read 2 more answers

Interpret (1,15)<br><br> interpret (4,60)

statuscvo [17]

Interpreting (1,15) with (4,60), found that (4,60) is 4 times of the coordinate (1,15).

<h3>What are coordinates?</h3>

A pair of numbers that describe the position of a point on a coordinate plane by using the horizontal and vertical distances from the two reference axes.

For interpreting we have to find the relation between (1,15) and (4,60)

We have, (1,15)

For x- coordinate 1, y- coordinate will be 15

Now,

let us take x- coordinate=2 then y- coordinate =30

again, if x-coordinate = 3 then y-coordinate = 45

lastly, if the x-coordinate =4 then y-coordinate = 60

That means (4,60) is the 4 times of the coordinate (1,15) i.e., the x-coordinate is 4 times as well as the y- coordinate is 4 times.

Learn more about coordinates here:

brainly.com/question/23450276

#SPJ1

4 0

1 year ago

Each person in a group of college students was identified by graduating year and asked when he or she preferred taking classes:

swat32

Answer:

<em>0.615</em>

Step-by-step explanation:

The frequency table is attached below.

We have to calculate, the probability that the student preferred morning classes given he or she is a junior.

i.e $P(\text{Morning }|\text{ Junior})$

We know that,

$P(A\ |\ B)=\dfrac{P(A\ \cap\ B)}{P(B)}$

So,

$P(\text{Morning }|\text{ Junior})=\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}$

Putting the values from the table,

$\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}=\dfrac{\frac{16}{143}}{\frac{26}{143}}=\dfrac{16}{26}=0.615$

3 0

3 years ago

Below is the five-number summary for 144 hikers who recently completed the John Muir Trail (JMT). The variable is the amount of

bearhunter [10]

Answer:

The IQR is given by:

$IQR = Q_3 -Q_1 = 28-18= 10$

If we want to find any possible outliers we can use the following formulas for the limits:

$Lower= Q_1 - 1.5 IQR$

$Upper= Q_3 + 1.5 IQR$

And if we find the lower limt we got:

$Lower= Q_1 - 1.5 IQR= 18-1.5*10 = 3$

$Upper= Q_3 + 1.5 IQR= 28+1.5*10= 43$

So then the left boundary for this case would be 3 days

Step-by-step explanation:

For this case we have the following 5 number summary from the data of 144 values:

Minimum: 9 days

Q1: 18 days

Median: 21 days

Q3: 28 days

Maximum: 56 days

The IQR is given by:

$IQR = Q_3 -Q_1 = 28-18= 10$

If we want to find any possible outliers we can use the following formulas for the limits:

$Lower= Q_1 - 1.5 IQR$

$Upper= Q_3 + 1.5 IQR$

And if we find the lower limt we got:

$Lower= Q_1 - 1.5 IQR= 18-1.5*10 = 3$

$Upper= Q_3 + 1.5 IQR= 28+1.5*10= 43$

So then the left boundary for this case would be 3 days

6 0

3 years ago