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3 years ago

What is the probability that the roll of a six-sided die will be either even or odd? A) 1 6 B) 1 4 C) 1 2 D) 1

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

8 0

The answer will most likely be D.) 1

Step2247 [10]3 years ago

6 0

Answer:

D) 1

Step-by-step explanation:

Every whole number (including the ones on a die) are either even or odd, nothing else. So it is guaranteed for the number to be even or odd.

You might be interested in

If I buy something for $25.40 with 15% tax, what would it be?

KonstantinChe [14]

25.40

divide by 10 for 10%: 2.54

divide it by 2: 1.27

add 2.54 and 1.27: 3.81

add the 2 numbers together: 25.40+3.81= $29.21

Your answer is $29.21

3 0

3 years ago

Determine the intercepts of the line.

iogann1982 [59]

Answer:

y-intercept: (0,5)

x-intercept: (5/2,0)

Step-by-step explanation:

$-4x+7=2y-3$

$-4(0)+7=2y-3$

$0+7=2y-3$

$7=2y-3$

$10=2y$

$5=y$ <-- y-intercept: (0,5)

$-4x+7=2y-3$

$-4x+7=2(0)-3$

$-4x+7=-3$

$-4x=-10$

$x=\frac{5}{2}$ <-- x-intercept: (5/2,0)

4 0

3 years ago

60 x 5? help plzzzzzzzzzzzzzzzz

fredd [130]

Answer:

300

Step-by-step explanation:

6 x 5= 30, so just add one zero to 6 and add a zero to 30

4 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

IRINA_888 [86]

Answer:

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = 0.02145 = 2.1 %

Step-by-step explanation:

Explanation:-

Given mean of the Population 'μ ' = 51 months

Standard deviation of the Population 'σ' = 11 months

Let 'X' be the random variable of Normal distribution

Let 'X' = 73

$Z = \frac{x-mean}{S.D} = \frac{73-51}{11} = 2$

Let 'X' = 84

$Z = \frac{84-51}{11} = 3$

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = P( 2 < Z < 3)

= P( Z<3) - P( Z <2)

= 0.5 + A(3) - ( 0.5 + A(2))

= A(3) - A( 2)

= 0.49865 - 0.4772 ( From Normal table)

= 0.02145

P( 73 < x < 84 ) = 0.02145

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = 0.02145 = 2.1 %

7 0

4 years ago

Given IQ scores are approximately normally distributed with a mean of 100 and a standard deviation of 15, the proportion of peop

weeeeeb [17]

The proportion of people with IQs above 130 is 2.3%.

<h3>How to calculate the value?</h3>

Normal distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

Here the mean is 100 and a standard deviation of 15.

µ = 100, σ = 15

P(X > 130) =

= P( (X-µ)/σ > (130-100)/15)

= P(z > 2)

= 1 - P(z < 2)

Using excel function:

= 1 - NORM.S.DIST(2, 1)

= 0.023 = 2.3%

The value is 2.3%.

Learn more about proportion on:

brainly.com/question/19994681

#SPJ1

8 0

2 years ago