Need help pleaseeeee!!

John has a die and a coin. The die is labeled 1 through 6 and the two sides of the coin are heads and tails. The tables below sh

Alenkasestr [34]

Answer:

n = 60

Step-by-step explanation:

Total outcomes, while tossing a coin are 2, given as:

{Heads, Tails}

The probability of getting a heads up, while tossing the coin is:

P(Heads) = Favorable Outcome/Total Outcomes

P(Heads) = 1/2 = 0.5

Now, the total outcomes, while rolling a die are 6, given as:

{1, 2, 3, 4, 5, 6}

5 of these numbers are factors of 12, which are:

{1, 2, 3, 4, 6}

Thus, the probability of getting a factor of 12, while rolling a die is:

P(Factor of 12) = Favorable Outcome/Total Outcomes

P(Factor of 12) = 5/6 = 0.83

So, the probability of getting both heads and factor of 12 will be:

P(Heads and Factor of 12) = P(Heads ∪ Factor of 12)

P(Heads and Factor of 12) = P(Heads) * P(Factor of 12) = 0.5 * 0.83

P(Heads and Factor of 12) = 0.417 = 41.7%

So, for 144 trials, the number of trials in which we get heads and factor of 12, are given by:

n = P(Heads and Factor of 12) * 144

where,

n = no. of trials John would expect to roll a number on the die that is a factor of 12 and toss a coin that lands heads-up out of next 144 trials

n = 0.417 * 144

n = 60

7 0

3 years ago

Is the function show liners or nonlinear?Explain

SpyIntel [72]

Non linear dneofnzkwodo

4 0

3 years ago

Read 2 more answers

Find tan 0. (Marking brainliest if correct) 8/17 15/17 15/8 8/15

Rina8888 [55]

Answer:

15/8

Step-by-step explanation:

Assuming you mean tan theta

Tan(theta) = opposite/adjacent

We know the adjacent side is 8 so the bottom of the fraction must also be 8

C: 15/8 is the only option that fits this criteria

4 0

2 years ago

the area of a rectangle is 10 square units the length is 2 units and the width is y units 10 = 2y what is the value of y. y = 12

frutty [35]

Answer:

2(5+y)+2y=38 10+2y+2y=38 4y=28 y=7 i think

Step-by-step explanation:

4 0

3 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the area of the region bounded by y = x³ and y = x is approximately 0.267 square units.

6 0

4 years ago