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FrozenT [24]
3 years ago
10

PLEASE HELP ME WITH MY MATH

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
7 0
8              2
16             4
20            5
28            7

ratio": number of bunch to litters of serving = 4:1 
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a baseball helmet cost $42 coach Rodger said he would pay $274 for 7 helmets is the exact answer reasonable
Zepler [3.9K]
No cause if I helmet cost $42 , 7 helmets will be $294.
$42 = 1 helmet
$? = 7 helmets
cross multiply = 42 × 7 = $294
7 0
3 years ago
<img src="https://tex.z-dn.net/?f=2%20%5Ctimes%202" id="TexFormula1" title="2 \times 2" alt="2 \times 2" align="absmiddle" class
EastWind [94]
4 (i literally had to think about this cause i thought it was too easy )
6 0
3 years ago
You are an assistant director of the alumni association at a local university. You attend a presentation given by the university
NNADVOKAT [17]

Answer:

(0.102, -0.062)

Step-by-step explanation:

sample size in 2018 = n1 = 216

sample size in 2017 = n2 = 200

number of people who went for another degree in 2018 = x1 = 54

number of people who went for another degree in 2017 = x2 = 46

p1 = x1/n1 = 0.25

p2 = x2/n2 = 0.23

At 95% confidence level, z critical = 1.96

now we have to solve for the confidence interval =

<h2> p1 -p2 ± z*\sqrt{((1-p1)*p1)/n1 + ((1-p2)*p2/n2}</h2>

0.25 -0.23 ± 1.96*\sqrt{((1 - 0.25) * 0.25)/216 + ((1 - 0.23) *0.23/200}

= 0.02 ± 1.96 * 0.042

= 0.02 + 0.082 = <u>0.102</u>

= 0.02 - 0.082 = <u>-0.062</u>

<u>There is 95% confidence that there is a difference that lies between  - 0.062 and 0.102 on the proportion of students who continued their education in the years, 2017 and 2018.</u>

<u></u>

<u>There is no significant difference between the two.</u>

5 0
3 years ago
It’s a new semester! Students are grouped into three clubs, which each has 10, 4 and 5 students. In how many ways can teacher se
ozzi

Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.

So there are 3 clubs:

  • Club A, with 10 students.
  • Club B, with 4 students.
  • Club C, with 5 students.

The possible combinations of 2 students from different clubs are

  • Club A with club B
  • Club A with club C
  • Club B with club C.

The number of combinations for each of these is given by the product between the number of students in the club, so we get:

  • Club A with club B: 10*4 = 40
  • Club A with club C: 10*5 = 50
  • Club B with club C. 4*5 = 20

For a total of 40 + 50 + 20 = 110 different combinations.

This means that there are 110 different ways in which 2 students from different clubs can be selected.

If you want to learn more about combination and selections, you can read:

brainly.com/question/251701

6 0
2 years ago
8-6x&gt;-18 solve for x
EleoNora [17]

Answer:

x=5

Step-by-step explanation:

Well all i did was guess a random number from 1-10 so I could see what numbers make the equation bigger than -18 and the fist thing I guessed was 5 so this is what I did

8-6*5=-22

And them I said let me see if any multiples equal -18 so I went to for and did this

8-6*4=-16

That's when I guessed the answer if i looked at the sign correctly!

8 0
2 years ago
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