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olya-2409 [2.1K]

3 years ago

7

Sally has a candy bar. if she gives 2/5th of the bar to mark and mark gives 1/2 of his piece to john, what percentage of the ori

ginal bar does john have?

1 answer:

liq [111]3 years ago

8 0

mark has 20% because half of 2/5 is 1/5 and 1/5 as a percent is 20%.

Hope this helped!

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Find the unit rate for miles per hour. Round to the nearest hundredth if necessary. A cyclist rides 65 miles in 3 hours.

Natasha2012 [34]

22 (rounded to the nearest one)

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3 years ago

The heights of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. Men the same

ohaa [14]

Answer:

z=(x-mu)/o

Step-by-step explanation:

X - Heights of women aged 20 to 29: X is N(64, 2.7)

Y - Heights of men aged 20 to 29: X is N(69.3, 2.8)

a) X=6'=72"

b) Y = 72"

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d) Y<5'5" =65"

e) X>73"

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3 0

2 years ago

What is the base of a triangle that has a height of 6 centimeters and an area of 18 centimeters? Use the formula h = StartFracti

azamat

Answer:

h = 6 cm

Step-by-step explanation:

Atriangle = b×h/2

18cm² = 6cm×h/2

36cm² = 6cm×h

h = 36cm²/6cm

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3 0

3 years ago

Read 2 more answers

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0

3 years ago

What's the difference between 1,968 and 3,000? Check your using inverse operations.

gtnhenbr [62]

The answer is 1,032. (also known as answer choice A)

6 0

3 years ago