All categories

3 years ago

Is 50% of 150 larger than 100

Mathematics

2 answers:

Doss [256]3 years ago

7 0

Answer:

Yes, half of 150 would be 75

Step-by-step explanation:

Half of 100 is 50, or 50% of 100 is 50, and half of 150 is 75, which could have been found by dividing 150 by 2.

RUDIKE [14]3 years ago

4 0

answer: no

because when you find the percent you will get 75 and 75 is not greater than 100 and im 100 percent correct

You might be interested in

Help ASAP!

Tema [17]

Answer:

Step-by-step explanation:

$\boxed{area \: of \: sector = \frac{1}{2} {r}^{2} θ }$

Radius= diameter ÷2

r= 32 ÷2

r= 16

$\frac{1}{2} {r}^{2} θ = \frac{512\pi}{3}$

Substitute r= 16,

$\frac{1}{2} ( {16}^{2} )(θ) = \frac{512\pi}{3}$

$128 \: θ = \frac{512\pi}{3} \\ θ = \frac{512\pi}{3} \div 128 \\ θ = \frac{512\pi}{384}$

Divide the denominator and numerator by 128:

$θ = \frac{4\pi}{3}$

Thus, the answer is B.

3 0

3 years ago

What is the reason for step 3 of this proof?

Reika [66]

Answer: D

Explanation:

It should be vertical angles theorem

8 0

3 years ago

What is the truth value for the following conditional statement?

alexgriva [62]

True I think is the answer

8 0

3 years ago

Read 2 more answers

Please help. This is a math vocabulary crossword puzzle

emmasim [6.3K]

Answer: i cant read that m8 take a better picture

Step-by-step explanation:

7 0

4 years ago

Do bonds reduce the overall risk of an investment portfolio? Let x be a random variable representing annual percent return for V

ryzh [129]

Answer:

∑x=109, ∑x²=4173, ∑y=87, ∑y²=2083.

sample mean for x=10.9.

variance for x=331.66.

standard deviation for x=18.21.

sample mean for y=8.7.

variance for y=147.34.

standard deviation for y=12.14.

75% Chebyshev interval around the mean for x values

-25.52,47.32

75% Chebyshev interval around the mean for y values

-15.58,32.98

The interval shows that Vanguard Balanced index has smaller spread.

coefficient of variation for x=167.06%

coefficient of variation for y=139.54%

Smaller CV means lower risk so, smaller CV is better.

Step-by-step explanation:

Note: ALL ANSWERS ARE ROUNDED TO TWO DECIMAL PLACES

∑x=17+0+17+28+28+27+29−12−12−13=109,

∑x²=17²+0²+17²+28²+28²+27²+29²+(−12)²+(−12)²+(−13)²=4173,

∑y=14−2+27+18+20+11+14−2−3−10=87,

∑y²=14²+(−2)²+(27)²+(18)²+(20)²+(11)²+(14)²(−2)²+(−3)²+(−10)²=2083.

sample mean for x=∑x/n=109/10=10.9.

$variance x=\frac{sum(x-mean )^2}{n-1}$

x-mean

6.10

-10.90

6.10

17.10

16.10

18.10

-22.90

-23.90

sum(x-mean)²=6.10 ²+(-10.90 )²+6.10 ²+17.10² +17.10²+ 16.10² +18.10² +(-22.90 )²+(-22.90 )²+(-23.90 )²=2984.9

$variance x=\frac{2984.9}{9}$

Variance for x=331.66.

standard deviation for x=√variance for x

standard deviation for x=√331.66

standard deviation for x=18.21.

sample mean for y=∑y/n=87/10=8.7.

$variance y=\frac{sum(y-mean )^2}{n-1}$

y-mean

5.30

-10.70

18.30

9.30

11.30

2.30

5.30

-10.70

-11.70

-18.70

sum(y-mean)²=5.30 ²+(-10.70 )²+18.30² +9.30² +11.30² +2.30² +5.30² +(-10.70 )²+(-11.70 )²+(-18.70 )²=1326.1

$variance y=\frac{1326.1}{9}$

variance for y=147.34.

standard deviation for y=√variance for y

standard deviation for y=√147.34

standard deviation for y=12.14.

75% Chebyshev Interval

mean±k standard deviation

We have to find value of k.

$1-\frac{1}{k^{2} } =0.75$

$1-0.75 =\frac{1}{k^{2} }$

$\frac{1}{k^{2} }=0.25$

k²=4

k=2

so, 75% Chebyshev Interval for x

10.9±2*(18.21)

[-25.52,47.32]

75% Chebyshev Interval for y

8.7±2*(12.14)

[-15.58,32.98]

The interval shows that Vanguard Balanced index has smaller spread.

$CV for x=\frac{S.Dx}{meanx} *100$

$CV for x=\frac{18.21}{10.9} *100$

CV for x=167.06%

CV for y

$CV for y=\frac{S.Dy}{meany} *100$

$CV for y=\frac{12.14}{8.7} *100$

Cv for y=139.54%

The smaller CV shows the smaller amount of risk so, smaller CV is better.

4 0

3 years ago