All categories

3 years ago

Which is bigger 12.5% or 3/20

Mathematics

1 answer:

DaniilM [7]3 years ago

4 0

It's 0.125 which is 12.5% or 0.15 which is 3/20 in fraction form so 3/20 is bigger

You might be interested in

Help this is high school mathematics and even though I’m usually good I need a little help it would be nice if you could also pr

Kitty [74]

Step-by-step explanation:

I'm not sure if this correct or not

7 0

3 years ago

Find the vertex, focus, directrix, and focal width of the parabola. x2 = 28y

katen-ka-za [31]

X^2 = 28y
(x - 0)^2 = 4*7 (y - 0) 
a = 7

5 0

3 years ago

Read 2 more answers

A survey of 400 randomly selected high school students determined that 68 play organized sports. (a) What is the probability th

Elina [12.6K]

Answer: a) The probability that a randomly selected high school student plays organized sports = 0.17

b) The randomly selected high school student is unlikely to play organized sports.

Step-by-step explanation:

Given : A survey of 400 randomly selected high school students determined that 68 play organized sports.

Then , the probability that a randomly selected high school student plays organized sports is given by :-

$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{68}{400}=0.17$

In percent , the probability that a randomly selected high school student plays organized sports is 17%.

since 17% lies in the interval of unlikely events (0%, 25%).

It means that the randomly selected high school student is unlikely to play organized sports.

3 0

3 years ago

PLEASE HELP ON MATH QUESTIONS!!!!!

agasfer [191]

This is not math

Explation:

5 0

3 years ago

City health officials will conduct a two-sample t-test for a difference in means to investigate whether

Anvisha [2.4K]

Answer:

The calculated value t = 2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means

Step-by-step explanation:

Step(i):-

Given that the first sample size 'n₁' = 22

Given that the mean of the first sample x₁⁻ = 120

Given that the standard deviation of the sample (s₁) = 20

Given that the mean of the second sample x₂⁻ = 100

Given that the standard deviation of the second sample (S₂) = 30

Step(ii):-

T-test statistic

$t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

where

$S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2} S_{2} ^{2} }{n_{1} +n_{2}-2 }$

$S^{2} = \frac{22( 20)^{2}+ 15(30) ^{2} }{22 +15-2 }$

S² = 637.1428

S = √637.1428 = 25.24168

The standard deviation of the sample 'S' = 25.24168

Step(iii):-

Null Hypothesis:H₀:

There is no difference between the means

Alternative Hypothesis:H₁:

There is a difference between the means

T-test statistic

$t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

$t = \frac{ 120 -100 }{\sqrt{637.14 (\frac{1}{22 }+\frac{1}{15 } ) } }$

t = 2.366

Degrees of freedom

γ = n₁+n₂ -2 = 22+15-2 = 35

t₀.₀₅ = 2.0301

The calculated value t = 2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means

3 0

3 years ago

Read 2 more answers