Solve for x please show work... i will give you brainliest

If Point P is between Q and R and QP = 16 and Qr = 52. What is the length of PR?

Dimas [21]

1) Here, QR = QP + PR

52 = 16 + PR

PR = 52 - 16 = 36

In short, Your Answer would be Option C

2) Slope of mentioned line:
m = (5+5) / (0-4)
m = 10/-4
m = -5/2

Parallel lines have same slope so it would be -5/2 as well

In short, Your Answer would be Option B

Hope this helps!

5 0

3 years ago

Statistics show that about 42% of Americans voted in the previous national election. If three Americans are randomly selected, w

MrRa [10]

Answer:

19.51% probability that none of them voted in the last election

Step-by-step explanation:

For each American, there are only two possible outcomes. Either they voted in the previous national election, or they did not. The probability of an American voting in the previous election is independent of other Americans. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

42% of Americans voted in the previous national election.

This means that $p = 0.42$

Three Americans are randomly selected

This means that $n = 3$

What is the probability that none of them voted in the last election

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.42)^{0}.(0.58)^{3} = 0.1951$

19.51% probability that none of them voted in the last election

6 0

3 years ago

Can you guy's please help me out I was absent and now I don't have a clue!

Elena-2011 [213]

For these problems you have to use PEMDAS so for problem 15 you would do the parentheses first, (21-3) you get 18 and then exponents, 3^2, you get 9, no multiplication but there is division so take your answers, 18 and 9 and divide them, you get 2, so the answer to 15 is 2.

6 0

3 years ago

Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams

Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0

3 years ago

A book has 55 pages with pictures and 45 pages without pictures, which ratio relationship is shown by the pages in the book?

suter [353]

Answer:

11 pages with picture to every nine pages without pictures. or 11 per every 9

Step-by-step explanation:

divide each number by a common denominator, in this case 5, to get a change from 55 per every 45 to 11 per every 9.

8 0

3 years ago