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4 years ago

What is the theoretical probability of rolling a 3?

Mathematics

1 answer:

aleksandrvk [35]4 years ago

7 0

Answer:

1/6

Step-by-step explanation:

the number on the bottom of the fraction (denominator) is equal to the total number of possibilities (in this case there are 6 possibilities). for the number on top (the numerator) we are trying to work out the probability of rolling a 3. a 3 is only 1 of the 6 options (1, 2, <u>3</u>, 4, 5, 6) so the number on top is 1.

I hope this was helpful :-)

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Given tantheta = 4/3 and pi < theta < 3pi/2; find cos2theta

Sidana [21]

Answer:

d. -7/25

Step-by-step explanation:

π<θ<3π/2 in third quadrant

tanθ = 4/3

sinθ = - 4/5

cosθ = - 3/5

cos 2θ = cos²θ - sin²θ = (- 3/5)² - (- 4/5)² = 9/25 - 16/25 = - 7/25

8 0

3 years ago

Read 2 more answers

Tell whether the equation has one, zero, or infinitely many solutions.

ahrayia [7]

Answer:

Infinitely many solutions.

Step-by-step explanation:

4(x-8)+9=4x-23

4x-32+9=4x-23

4x-23=4x-23

infinitely many solutions.

7 0

3 years ago

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$