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kicyunya [14]
3 years ago
14

7. A hotelier conducted a survey of guests staying at her hotel. The table shows some

Mathematics
1 answer:
jonny [76]3 years ago
8 0

Answer:

here's the solution :-

Step-by-step explanation:

i hope it helped...

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a mixture of medicine consists of three vitamins A, B and C,if the ratio of the vitamins should be 5:1:3 and the total mixture i
Sergio039 [100]

Answer:

A≈22 | B≈4.4 | C≈13.2

Step-by-step explanation:

5 + 1 + 3= 9

40/9≈ 4.444…

(4.4x5=22)+(1x4.4=4.4)+(3x4=13.2)= 39.6≈40


Idrk  if I did this right, but I hope it’s kinda useful (I have a feeling I misunderstood the question lmdao)

8 0
3 years ago
What re the answers to 20-30
BARSIC [14]

Answer:

20.

m =  -52

22.

x=13

24.

a=-90

26.

n=-2

28.

a=9

30.

m=-10

Step-by-step explanation:

20. m/4=-13

m = -13*4 = -52

22. -143=-11x

x=-143/-11 =13

24. -5=a/18

a=-5*18= -90

26. n-8=-10

n=--10+8

n=-2

28. a+11=20

a=20-11

a=9

30. 18+m=8

m=8-18

m=-10

7 0
3 years ago
Calculating percentages. no links
Alenkasestr [34]

Answer:

35%

Step-by-step explanation:

Its 7/20 so 0.35 and then times 100

4 0
3 years ago
Read 2 more answers
You are buying lawn seed for the the football field a 10 pound bag of lawn seed covers 2000 square feet.
Masja [62]

A football field is 120 yards long (100 yards playing field plus 2 end zones 10 yards each) by 53 1/3 yards wide.

1 yard = 3 feet.

120 yards x 3 feet per yard = 360 feet

53 1/3 yards x 3 = 160 feet.

The area of the football fields is 360 x 160 = 57,600 square feet.

1 bag covers 2000 square feet:

57,600 / 2,000 = 28.8 bags, round up to 29 bags are needed.

1 bag cost $27.79.

29 bags x $27.79 = $805.91 total cost.

7 0
3 years ago
Read 2 more answers
A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard
bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

  = \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2

value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

8 0
3 years ago
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