Which value of xis

J is the midpoint of HK¯¯¯¯¯¯¯ . What are HJ, JK, and HK?

lana [24]

Since J is the midpoint of HK, that means HK is split into two sections HJ and JK that are the same length.

1) You are told that the measure of segment HJ = 9x-2 and that of segment JK = 4x+13. Since you also know they are equal lengths, you can set these equations equal to each other to find the value of x!
HJ = JK
9x-2 = 4x+13
5x = 15
x = 3

2) Now you know x = 3. Plug that into your given equations for HJ and JK to find the length of each segment (or a shortcut would be to find one of them, and then you also know the other is the same length. I'm doing both, just to make sure I don't make a silly mistake!):
HJ = 9x-2
HJ = 9(3) - 2
HJ = 27 - 2
HJ = 25

JK = 4x + 13
JK = 4(3) + 13
JK = 12 + 13
JK = 25

3) Finally, the length of HK is just the length of HJ + JK, or HK = 25 + 25 = 50.

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Answer: HJ = 25, JK = 25, HK = 50

4 0

3 years ago

Help me please urgent

asambeis [7]

$Let \: x \: be \: the \: number \: of \: hours \\ Expession: - 4.5x = - 27\Leftrightarrow x = \frac{ - 27}{ - 4.5} = 6 \: hours$

7 0

2 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

3 years ago

Um can some one pls explain this

slamgirl [31]

Answer:

the answer is c) 182.25

Step-by-step explanation:

The information provided gives you a scale that will ultimately allow you to determine the height of the larger cylinder.

Since it tells us the two are similar, we can use the scale to solve for the second height.

Set up your math like this:

$\frac{16}{h}=\frac{64}{729}$

h represents the height you are trying to find.

using a common multiple, you can solve for h.

16 times 4 is 64. so 729 divided by 4 is 182.25

5 0

3 years ago

Read 2 more answers

Answer this question and tell me what you get

Vlada [557]

Option B is the answer because angle A is greater than C and A

7 0

2 years ago