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3 years ago

Help please! asap! Thank you!

Mathematics

1 answer:

Nady [450]3 years ago

7 0

Answer:

A and E

Step-by-step explanation:

A will give the total pound.

E will give the number of servings.

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According to a survey of 100 people regarding their movie attendance in the last year, 40 had seen a science fiction movie, 55 h

AnnyKZ [126]

Answer:

Check attached picture ; 75 ; 40 ; 25%

Step-by-step explanation:

s = science fiction movie

H = horror movie

A = adventure movie

From the information given :

n(SnHnA) = 5

n(SnH) only = 15 - 5 = 10

n(SnA) only = 25 - 5 = 20

n(HnA) only = 5 - 5 = 0

n(S) only = 40 - (10+5+20) = 5

n(H) only = 20 - (10+5+0) = 5

n(A) only = 55 - (5 + 20 + 0) = 30

b.) Number of people who saw atleast one movie :

Sum of all numbers in the circles :

(5 + 20 + 5 + 10 + 5 + 30 + 0) = 75

c. How many people saw only one type of movie?

n(S) only + n(H) only + n(A) only

(5 + 5 + 30) = 40

d. What percentage of the people saw none of the three movie types?

Number of people who saw none of the movies:

100 - (Number of people who saw atleast one)

100 - 75 = 25

Hence, percentage who saw none :

(25/ 100) * 100%

(1/4) * 100%

= 25%

3 0

3 years ago

Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. The perpe

lubasha [3.4K]

Answer:

(a)In the attachment

(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.

Step-by-step explanation:

(a)In the attachment

(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)

First, let us determine length of line RM.

Using Pythagoras theorem

$|RM|^{2}=30^2+x^2\\|RM|=\sqrt{30^2+x^2}$

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.

Time Taken = Distance/Time

Time taken on the road $=\frac{\sqrt{30^2+x^2}}{60}$

Time taken on the highway $=\frac{50-x}{110}$

Total time taken to travel, T $=\frac{\sqrt{30^2+x^2}}{60}+\frac{50-x}{110}$

Minimum time taken occurs when the derivative of T equals 0.

$T^{'}=\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}\\\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}=0\\\frac{x}{60\sqrt{30^2+x^2}}=\frac{1}{110}\\110x=60\sqrt{30^2+x^2}\\$

Square both sides

$12100x^2=3600(30^2+x^2)\\12100x^2=3240000+3600x^2\\12100x^2-3600x^2=3240000\\8500x^2=3240000\\x^2=\frac{3240000}{8500} =381.18\\x=\sqrt{381.18} =19.52$