Ask question

Ask question

All categories

3 years ago

14

A number is selected at random from the numbers 1,4,9,16 and another number y from 1,2,3,4 find the probability that the value o

f xy is more than 16

2 answers:

svetoff [14.1K]3 years ago

8 0

It have to be 6/8 or 3/4 cause all of them are over 16

kirill [66]3 years ago

5 0

Lets draw a sample space of the possible outcomes of xy-

S= (1, 2, 3, 4, 4, 8, 12, 16, 9, 18, 27, 36, 16, 32, 48, 64)

Total no. of outcomes= 4*4= 16

No. of outcomes greater than than 16= 6

So; Probability of xy>16 = 6/16 = 3/8

You might be interested in

The range of y= square root x-5-1

Airida [17]

Answer:

^x-6

Step-by-step explanation:

6 0

2 years ago

Suppose 42% of the population has myopia. If a random sample of size 442 is selected, what is the probability that the proportio

sineoko [7]

Answer:

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose 42% of the population has myopia.

This means that $p = 0.42$

Random sample of size 442 is selected

This means that $n = 442$

Mean and standard deviation:

$\mu = p = 0.42$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42*0.58}{442}} = 0.0235$

What is the probability that the proportion of persons with myopia will differ from the population proportion by less than 3%?

Proportion between 0.42 + 0.03 = 0.45 and 0.42 - 0.03 = 0.39, which is the p-value of Z when X = 0.45 subtracted by the p-value of Z when X = 0.39.

X = 0.45

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.45 - 0.42}{0.0235}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

X = 0.39

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.39 - 0.42}{0.0235}$

$Z = -1.28$

$Z = -1.28$ has a p-value of 0.1003

0.8997 - 0.1003 = 0.7994

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.

7 0

3 years ago

In a trivia contest, players from teams and work together to earn as many points as possible for their team. Each team can have

seropon [69]

Answer:

The inequality for each quantity described is given as follows;

0 ≤ A + B + C + D + E ≤ 50

Step-by-step explanation:

The given information are;

The number of players each team can have = between 3 and 5

The maximum number of points a player can score in each round of the game = 10 points

The number of players in Elena's team = Elena + 4 = 5 players

The total number of points Elena's team earns at the end of the round is given as follows;

0 ≤ A + B + C + D + E ≤ 5 × 10

Where the variables A, B, C, D, and E are the points each of Elena and are makes such that the minimum points is 0 + 0 + 0 + 0 + 0 = 0 and the maximum point is 10 + 10 + 10 + 10 + 10 = 5 × 10 = 50, which gives;

0 ≤ A + B + C + D + E ≤ 50.

8 0

3 years ago

For each x and n, find the multiplicative inverse mod n of x. Your answer should be an integer s in the range 0 through n - 1. C

krok68 [10]

Use the Euclidean algorithm to express 1 as a linear combination of $x$ and $n$ .

a. $52^{-1}\equiv40\pmod{77}$ because

77 = 1*52 + 25

52 = 2*25 + 2

25 = 12*2 + 1

so we can write

1 = 25 - 12*2 = 25*25 - 12*52 = (77 - 52)(77 - 52) - 12*52 = 77^2 - 2*52*77 + 52^2 - 12*52

Taken modulo 77 leaves us with

$1\equiv52\cdot52-12\cdot52\equiv40\cdot52\pmod{77}\implies52^{-1}\equiv40\pmod{77}$

b. First, $77\equiv25\pmod{52}$ , so really we're looking for the inverse of 25 mod 52. We've basically done the work in part (a) already:

1 = 25*25 - 12*52

Taken modulo 52, we're left with

$1\equiv25\cdot25\pmod{52}\implies25^{-1}\equiv25\pmod{52}$

c. The EA gives

71 = 1*53 + 18

53 = 2*18 + 17

18 = 1*17 + 1

so we get

1 = 18 - 17 = 3*18 - 53 = 3*71 - 4*53

so that taken module 71, we find

$1\equiv(-4)\cdot53\pmod{71}\implies53^{-1}\equiv-4\equiv67\pmod{71}$

d. Same process as with (b). First we have $71\equiv18\pmod{53}$ , and we've already shown that

1 = 3*18 - 53

which means, taken modulo 53, that

$1\equiv3\cdot18\pmod{53}\implies71^{-1}\equiv18^{-1}\equiv3\pmod{53}$

6 0

3 years ago

Number 20. if a and b are integers and a > b, then l a l > l b l

raketka [301]

. Answer: I am pretty sure sometimes

Step-by-step explanation: if a = 6 and B = - 7, if you put them l a l > | b |

It makes them both positive so sometimes is the most probable answer .

( Sorry if it’s wrong)

6 0

3 years ago

Read 2 more answers