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3 years ago

15

A coin is tossed and an eight-sided die numbered 1 through 8 is rolled. Find the probability of tossing a head and then rolling

a number greater than 5.

1 answer:

ratelena [41]3 years ago

6 0

Answer: 7/8

Step-by-step explanation:

Probability is calculated by number of possible outcomes/ total outcomes

From the question, a coin only has two sides, Head or Tail. Total outcomes is therefore equal to 2

Possible outcome is to toss a head, which is one event

Probability of tossing a head is therefore =1/2

The second part of the question is rolling a die with eight sides.

Here, the die is said to have eight sides, 1,2,3,4,5,6,7,8. Total outcomes is therefore equal to 8

Possible outcome of rolling a number greater than 5 is 3 possibilities, that is rolling 6, 7 and 8.

Pr(rolling > 5) is 3/8

Therefore, Pr ( head; rolling > than 5)

= 1/2 + 3/8

=7/8

I hope this is clear, please mark as brainliest answer

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43% of a sample of 100 visitors to a national park believes that additional funds should be used to preserve endangered species.

IrinaVladis [17]

Answer:

A. 9.7%

Step-by-step explanation:

margin of error:

z * $\sqrt{\frac{p(1-p)}{N} }$

n is sample size

z is the z score (for a 95% confidence interval you would use 1.96)

p is the sample proportion

to solve for 'p', remember that if a decimal value is given, use that. but if it is a whole number, then divide that number by the sample size, and substitute it into the formula.

in this case we are given 43% so 0.43.

in the formula :

1.96 * $\sqrt{\frac{0.43(1-0.43)}{100} }$

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2 years ago

Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

5 0

3 years ago

Can you answer this math homework? Please!

luda_lava [24]

Answer: (3.5,-2)

Step-by-step explanation:

To solve this problem, all we have to do is plug in y=-2 into either equations from step one to find the coordinate. To make sure we have the right x-coordinate, let's plug in y for both equations.

4x-7y=28 [plug in y=-2]

4x-7(-2)=28 [multiply]

4x+14=28 [subtract both sides by 14]

4x=14 [divide both sides by 4]

x=3.5

------------------------------------------------------------------------------------------------------------

6x-5y=31 [plug in y=-2]

6x-5(-2)=31 [multiply]

6x+10=31 [subtract both sides by 10]

6x=21 [divide both sides by 6]

x=3.5

Now, we know the solution to the system is (3.5,-2).

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2 years ago

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katovenus [111]

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