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3 years ago

4 - (-18) = 4 - (-18) =

Mathematics

1 answer:

VashaNatasha [74]3 years ago

8 0

Answer: 22

Step-by-step explanation:

4 - (-18) = 4 + (+ 18) = 22

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From the table below create an equation.

Bond [772]

Answer:

Step-by-step explanation:

2,4 ,10,12,19,36.5,40

4 0

3 years ago

A study claims that the mean age of online dating service users is 40 years. Some researchers think this is not accurate and wan

Mademuasel [1]

Answer: $H_0:\mu=40$

$H_0:\neq40$

Step-by-step explanation:

A null hypothesis is a hypothesis where a researcher generally try to disprove, it says that there is no statistically significant relationship between the two variables .
An alternative hypothesis says that there is a statistical significance between two variables.

Claim 1. : Mean age of online dating service users is 40 years.

i.e. $\mu=40$ , since it has equals sign so we take this as null hypothesis.

Claim 2. : Mean age of online dating service users is not 40 years.

$\mu\neq40$

⇒ Null Hypothesis : $H_0:\mu=40$

Alternative hypothesis : $H_0:\neq40$

5 0

3 years ago

Which describes how to find the solution to 5k>22 ?

alex41 [277]

You need k by itself, so the only way to do that is to divide both sides by 5

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3 years ago

Jasmine cell phone company charges her $35 a month for phone service plus $.50 for each text message how many text messages does

aksik [14]

She would be able to send 34 messages

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3 years ago

There are three clubs at Redwood High School: Debate, Student council, and Key club. There are 1000 students in the school. And

hram777 [196]

Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 - 170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0

3 years ago