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3 years ago

You roll a number cube numbered form 1 to 6 what is the probability that the number is divisible by 2

Mathematics

2 answers:

inysia [295]3 years ago

6 0

Answer:

The probability is 1/2 or 50%

Step-by-step explanation:

We have 3 numbers that are divisible by 2 (2, 4, 6)

We find the probability by dividing the favorable results (3) by the # of all results (6).

3/6=1/2=0.5=50%

Enjoy! -Dante G. :)

aleksley [76]3 years ago

3 0

Answer:

1/2

Step-by-step explanation:

Probability is essentially (# times a specific event occurs) / (# times the general event occurs).

Here, the specific event is rolling a number that is divisible by 2. The numbers divisible by 2 from 1 to 6 are 2, 4, and 6. Since there are 3 such numbers, the specific event can occur 3 times.

The general event is simply rolling any number. Because there are 6 numbers we can roll, the number of times the general event can occur is 6.

So, the probability is 3/6 = 1/2.

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3 years ago

Read 2 more answers

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)

Anna35 [415]

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is $f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}$ and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

f(a)=f(b)
f(x) must be continuous in [a,b].
f(x) must be differentiable in (a,b).

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

$\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end$

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

Learn more about Rolle's Theorem from here brainly.com/question/12279222

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8 0

2 years ago

I need help plsss!!!!!!!!!

Zepler [3.9K]

Answer:

kne-hnfd-vsq

Step-by-step explanation:

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5 0

3 years ago

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of sc

Shkiper50 [21]

Answer:

a) Scores of 2 and higher are significantly high

b) Scores of -2 and lower are significantly low

c) Scores between -2 and 2 are not significant.

Step-by-step explanation:

Mean = 0

Standard deviation = 1

a. significantly high (or at least 2 standard deviations above the mean).

2 standard deviations above the mean is:

0 + 1*2 = 2

So scores of 2 and higher are significantly high

b. significantly low (or at least 2 standard deviations below the mean).

2 standard deviations below the mean is:

0 - 1*2 = -2

So scores of -2 and lower are significantly low

c. not significant (or less than 2 standard deviations away from the mean).

2 standard deviations above the mean is:

0 + 1*2 = 2

2 standard deviations below the mean is:

0 - 1*2 = -2

So scores between -2 and 2 are not significant.

3 0

3 years ago

91.55

cricket20 [7]

Answer:

the answer is 3x2-10x+6=0

4 0

3 years ago