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3 years ago

Missy rolls two 6 sided number cubes labeled 1-6. What is the probability that both number cubes land on a prime number

Mathematics

1 answer:

WITCHER [35]3 years ago

8 0

Answer:

There is a 1/4 chance of Missy rolling a prime number on both cubes.

Step-by-step explanation:

Rolling one 6-sided cube, Missy has a 3/6 or 1/2 chance of landing on a prime number because the prime numbers between 1 and 6 (inclusive) are 2,3 and 5. When Missy rolls 2, 6-sided cubes, however, the chances of that happening lessens, and we multiply 1/2 by 1/2 as a result. Therefore, when Missy rolls two 6-sided cubes, there is a 1/4 chance of Missy rolling a prime number on both cubes.

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Colton gave out a survey to some students in his school about their favorite color. 470 of those surveyed said their favorite co

Tanzania [10]

Answer:

500

Step-by-step explanation:

Convert percent to decimal:

94%=

100

=0.94

Move decimal 2 places to the left

Multiplying the decimal by the whole gives us the part:

0.94x=

470

0.94

0.94x = 0.94

470

Divide both sides by 0.94x=

500

7 0

2 years ago

write an equation of the line that passes through the given point and has the given slope slope is -3 (-5, 4)

Paraphin [41]

point slope form of a line

y-y1 = m(x-x1)

y-4 = -3(x--5)

y-4 =-3(x+5)

to change to slope intercept form

dstribute

y-4 = -3x-15

add 4 to each side

y = -3x-15+4

y = -3x-11

3 0

3 years ago

Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

antoniya [11.8K]

The perimeter of the quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3) is 20units.

Option C) is the correct answer.

<h3>What a quadrilateral?</h3>

A quadrilateral is simply a polygon with four sides, four angles, and four vertices.

To get the perimeter, we simply add the values of the four side.

Given that;

The vertices are at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

To get the dimension between the given coordinates, we use;

d = √((x2 -x1)² +(y2 - y1)²)

For length CD, DE, EF and FC

CD = √((2 - (-2))² + (4 - 1)²) = √( 16+9) = √25 = 5

DE = √((5 - 2)² + (0 - 4)²) = √( 9+16) = √25 = 5

EF = √((1 - 5)² + (-3 - 0)²) = √( 16+9) = √25 = 5

FC = √((-2 - 1)² + ( 1 - (-3))²) = √( 9+16) = √25 = 5

Perimeter of the quadrilateral = CD + DE + EF + FC

Perimeter of the quadrilateral = 5 + 5 + 5 + 5

Perimeter of the quadrilateral = 20units

Therefore, the perimeter of the quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3) is 20units.

Option C) is the correct answer.

Learn more about area of rectangle here: brainly.com/question/27612962

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6 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP IM TAKING YHE IREADY TEST RN ... and I have more question so please help me answer them

Eduardwww [97]

Answer:

see the pic below.

3 0

3 years ago

An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr

Veseljchak [2.6K]

Answer:

The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance

Null hypothesis is rejected 

Alternative hypothesis is accepted

The extra carbonation of cola results in a higher average compression strength

Step-by-step explanation:

Step(i):-

Given data

First sample size (n₁) = 10

mean of the first sample(x₁⁻) = 537

standard deviation of the first sample (S₁) = 22

second sample size (n₂) = 10

mean of the second sample (x₂⁻) = 559

standard deviation of the second sample (S₂) = 17

Step(ii):-

Null hypothesis : H₀:- The extra carbonation of cola results in a lower average compression strength

Alternative Hypothesis :H₁

The extra carbonation of cola results in a higher average compression strength

Step(iii):-

By using student's t -test for difference of means

Test statistic

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

 where 

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

  $S^{2} = \frac{10(22)^{2} + 10 (17) ^{2} }{10+10 -2 } = \frac{ 7730}{18} = 429.4$ 

  $t = \frac{537-559 }{\sqrt{429.4 (\frac{1}{10 }+\frac{1}{10 } } )}$ 

 t = -2.375

|t| = |-2.375| = 2.375

Degrees of freedom

γ = n₁+n₂ -2 = 10+10-2 =18

 $t_{\frac{\alpha }{2} ,n-1}=t_{(\frac{0.05}{2} ,18)} = t_{(0.025,18)}}=2.1009$ 

The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance

Null hypothesis is rejected 

Alternative hypothesis is accepted

Final answer:-

The extra carbonation of cola results in a higher average compression strength

 

5 0

2 years ago