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MrMuchimi
1 year ago
13

Christian randomly selects students from his grade to rate a math test as easy, moderate, or difficult. Of the students he surve

yed, 13 said the test was easy, 11 rated it as moderate, and 3 found it difficult. Assuming that all students took the same test, how many of the 162 total students in Christian’s grade would probably rate the test something other than easy?
Mathematics
1 answer:
IRINA_888 [86]1 year ago
4 0

Answer:

84

Step-by-step explanation:

Total students he surveyed: 27

Amount that found it something OTHER than easy: 14 (11 moderate + 3 difficult)

This means that \frac{14}{27} found it something other than easy

Multiply that by 162 to find the proportion for the larger population:

162 * (14/27), and we get 84.

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A florist has to choose four different types of flowers to include in a bouquet. In how many ways can the florist do this, if th
Nat2105 [25]

Answer:

Step-by-step explanation:

Given:

Type of Flowers = 5

To choose = 4

Required

Number of ways 4 can be chosen

The first flower can be chosen in 5 ways

The second flower can be chosen in 4 ways

The third flower can be chosen in 3 ways

The fourth flower can be chosen in 2 ways

Total Number of Selection = 5 * 4 * 3 * 2

Total Number of Selection = 120 ways;

Alternatively, this can be solved using concept of Permutation;

Given that 4 flowers to be chosen from 5,

then n = 5 and r = 4

Such that

nPr = \frac{n!}{(n - r)!}

Substitute 5 for n and 4 for r

5P4 = \frac{5!}{(5 - 4)!}

5P4 = \frac{5!}{1!}

5P4 = \frac{5*4*3*2*1}{1}

5P4 = \frac{120}{1}

5P4 = 120

Hence, the number of ways the florist can chose 4 flowers from 5 is 120 ways

4 0
3 years ago
When Marco swim, the number of calories burned after time x, In minutes, is given by the equation
Cloud [144]
Answer : 4. 576


Y= 330 + 8.2(30)

330 +246

576




4 0
2 years ago
FUNDRAISING A school is raising money by selling calendars for $20 each. Mrs. Hawkins promised a party to whichever of her Engli
____ [38]

Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

,

• 2nd period class sold 123 calendars in total

We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

,

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

,

• 4th period class sold 126 calendars in total

Therefore, the average is:

• 3rd period class sold 89/4 = 22.25 calendars per week

,

• 4th period class sold 126/4 = 31.5 calendars per week

Summarizing, we have:

15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

•

7 0
1 year ago
Help pls with this trigonometry question
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7 0
2 years ago
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(6x3)2 = 62(x3)2 = 36x6<br> True or false
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Answer:

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Step-by-step explanation:

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