This is my last question please help

What is the probability of flipping a coin 10 times and getting heads four times Rania answer to the nearest 10th of a percent

EleoNora [17]

Answer: 0.205

Step-by-step explanation:

The probability in one flip is equal to 0.5.

Now, if we do 10 flips, we want 4 times that the coin ends in heads and the other 6 times it must end in tails, so the probability is:

0.5^4*0.5^6 = 0.5^10

Now, this is the case where in the first 4 flips we get heads and in the other 6 we get tails, but we have other ways to get only 4 heads (for example in the last 4 flips, 2 at the beginning and 2 at the end, etc)

So we also need to calculate all the possible permutations of 4 heads in 10 flips. this is:

C = 10!/(10 - 4)!*4! = (10*9*8*7)/(4*3*2*1) = 210

So the probability that we are looking for is:

P = (0.5^10)*210 = 0.205

4 0

4 years ago

Fix the inequality below to correctly show the domain of<br> the function.

Nataliya [291]

Answer:

from questions it is given that X is lies between the -9 and -5. In this question X lies only between -9 to -5 but -9 and -5 are excluded because there is no equal sign given in the question. so value of X are (-8,-7,-6). and plot in the graph .and these values are the component of X so they are also a domain.

6 0

2 years ago

84 is 12%of what number <br> please hurry!!!!!!

maks197457 [2]

Answer: 700

84 is 12% of What Number:

12% of 700 is 84

100% of 700 is 700, therefore 12 percent of 700 equals 84.

To check this, 12% of 700 IS 84 (Check on Calculator)

Hope this helps!!

5 0

3 years ago

WHATS THE ANSWER?? answer ASAP NO FILE AT ALL

Zepler [3.9K]

Answer:

A

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 71.5 m

Zinaida [17]

Answer:

a

$G = 0.007523$

b

The number of cars the highway patrol officer would watch before a car that is seen is $E(X) = 1.6027$

The standard deviation is $s = 0.9829$

gg

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 71.5 \ miles/hour$

The standard deviation is $\sigma = 4.75 \ miles/hour$

The speed limit is $x = 70 \ miles /hour$

Generally the probability of getting a car that is moving with speed greater than the speed limit is mathematically represented as

$p =P(X > x ) = P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > -0.31579 )$

Here

$\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of X )$

So

=> $p= P(X > 70) = P(Z > -0.31579 )$

From the z-table

$p = P(Z > -0.31579 ) = 0.62392$

So

$p = P(X > 70) = 0.62392$

Generally the probability of getting a car that is not moving with speed greater than the speed limit is mathematically represented as

$q = 1 - p$

=> $q = 1 - 0.62392$

=> $q = 0.37608$

Generally the probability of getting 5 cars that are not speeding is mathematically represented as

$G = q^5$

=> $G = (0.37608)^5$

=> $G = 0.007523$

Generally the number of cars that the highway patrol officer is expected to watch until the first car that is speeding is gotten is mathematically represented as

$E(X) = \frac{1}{p}$