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1 year ago

Show one way to count from $82 to $512

1 answer:

5 0

First way:
The easiest and simplest way is to count by 1 starting from 82 till you reach 512.
This will go as follows:
82, 83, 84, 85, ........... , 510, 511, 512

Second way:
We can note that the two given numbers are even numbers. This means that the two numbers are divisible by 2.
Therefore, we can count by 2 starting from 82 till we reach 512.
This will go as follows:
82, 84, 86, 88, ................... , 508, 510, 512

Third way:
We can note that the units digit in both numbers is the same (the digit is 2). This means that we can count from 82 till 512 by adding 10 each time.
This will go as follows:
82, 92, 102, 112, ......................, 492, 502, 512

Hope this helps :)

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Distinguish between rote and rational counting

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Answer: There is a difference between rote counting and rational counting. Rote counting involves the memorization of numbers. Rational counting tells children "how many there are." For children to count rationally, they need to demonstrate one-to-one correspondence.

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1 year ago

An airliner carries 150 passengers and has doors with a height of 78 in. Heights of men are normally distributed with a mean of

iogann1982 [59]

Answer:

Step-by-step explanation:

x, height of men is N(69, 2.8)

Sample size n =150

Hence sample std dev = $\frac{2.8}{\sqrt{150} } =0.229$

Hence Z score = $\frac{x-69}{0.229}$

A) Prob that a random man from 150 can fit without bending

= P(X<78) = P(Z<3.214)=1.0000

B) n =75

Sample std dev = $\frac{2.8}{\sqrt{75} } =0.3233$

P(X bar <72) = P(Z<9.28) = 1.00

C) Prob of B is more relevent because average male passengers would be more relevant than a single person

(D) The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.

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2 years ago

Two hockey teams, the Winnipeg Jets (yay!) and the Toronto Maple Leafs (boo!), miraculously make it to

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Answer:

There is a 54% chance that the Jets will win the Stanley Cup.

Step-by-step explanation:

Since two hockey teams, the Winnipeg Jets and the Toronto Maple Leafs, miraculously make it to the Stanley Cup finals, where as soon as a team wins 3 games, the championship is over, and the schedule for the playoffs for home games is: Jets - Leafs - Jets - Leafs - Jets; since if the Jets are playing at home, there is a 60% chance they'll win, while if they are playing on the road, there is a 45% chance they'll win, to find the probability that the Jets win the Stanley Cup the following calculation must be performed:

(0.60 + 0.45 + 0.60 + 0.45 + 0.60) / 5 = X

2.70 / 5 = X

0.54 = X

Thus, there is a 54% chance that the Jets will win the Stanley Cup.

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2 years ago

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2 years ago

Read 2 more answers

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

ASHA 777 [7]

Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem we have that:

Google Chrome has a 20.37% share of the browser market. This means that $p = 0.2037$

20 Internet users are sampled, so $n = 20$ .

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

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

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050$

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

$E(X) = np$

$E(X) = 20*0.2037 = 4.074$

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

$Var(X) = np(1-p)$

$Var(X) = 20*0.2037*(0.7963) = 3.2441$

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

$\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011$

The standard deviation for the number of Chrome users is 1.8011.

6 0

2 years ago