Which ordered pair makes both inequalities true?<br><br> y x

PLEASE help me. I am stuck on this question for a long time.

maria [59]

Answer: Plot 1

Step-by-step explanation:

#1 Find the median of the whole set

- since there are 31 data, it is odd, we just need to find the 16th data which is 73.

- the median is 73

#2 Find the median of Less than Q2

- since there are 15 data, it is odd, we just need to find the 8th data which is 64

- the median is 64

#3 Find the median of more than Q2

- since there are 15 data, it is odd, we just need to find the 8th data which is 84

- the median is 84.

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Putting it back to the box and whisker plot, it will be plot 1

6 0

3 years ago

An administrator surveys a random sample of 48 out of 900 middle school

Verizon [17]

Considering the definition of probability, 12 of the 48 students surveyed were in favor of the new dress code.

<h3>Definition of probability</h3>

Probability is the greater or lesser chance that a given event will occur.

In other words, the probability establishes a relationship between the number of favorable events and the total number of possible events.

Then, the probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases. This is called Laplace's Law.

P(A)=number of favorable events÷ number of total events

<h3>Probability that students are in favor of the new dress code</h3>

In this case, you know:

Total number of middle school students = 900 (number of possible cases)
The number of students are in favor of the new dress code = 225 (number of favorable cases)

Replacing in the definition of probability:

P(A)=225 students÷ 900 students

Solving:

Expressed as a percentage:

<h3>Number of the 48 students surveyed were that in favor of the new dress code</h3>

In this case, you know:

Total number of middle school students = 48 (number of possible cases)
25% students are in favor of the new dress code (P(A)= 25%= 0.25)

Replacing in the definition of probability:

0.25=students in favor of the new dress code÷ 48 students

Solving:

students in favor of the new dress code= 0.25×48 students

<u><em>students in favor of the new dress code= 12 students</em></u>

Finally, 12 of the 48 students surveyed were in favor of the new dress code.

Learn more about probability:

brainly.com/question/3283947

#SPJ1

3 0

2 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

Read 2 more answers

Scarlett is trying to find the height of a radio antenna on the roof of a local building. She stands at a horizontal distance of

Katarina [22]

The radio antennas' height is $4.6m$

From the attached diagram, the distance between point $AA$ and point $BB$ distance $H-h$ .

<h3 /><h3>Steps on how to determine the radio antennas' height</h3>

To find the distance $H$ , we make use of the relationship

$\dfrac{H}{33m}=tan33^\circ\\\\
\implies H=33m\cdot tan33^\circ$

To find the distance $h$ , we make use of the relationship

$\dfrac{h}{33m}=tan27^\circ\\\\
\implies h=33m\cdot tan27^\circ$

Thus, the distance $H-h$ , or the height of the radio antenna, can be found as follows

$H-h=33m\cdot tan33^\circ-33m\cdot tan27^\circ\\
=33m\cdot(tan33^\circ-tan27^\circ)\\
=33m\cdot(0.6494-0.5095)\\
=33m\cdot 0.1399\\
=4.6167m \approx 4.6m$