Write 5/6% as a decimal number

Write prime factorizations of the number 18.

Nat2105 [25]

Answer:

18 is a composite number. 18 = 1 x 18, 2 x 9, or 3 x 6. Factors of 18: 1, 2, 3, 6, 9, 18.

Prime factorization: 18 = 2 x 3 x 3, which can also be written 18 = 2 x 3²

hope the answer helped:)

4 0

3 years ago

Cole is an urban planner. He wants to create a small scale drawing of a city block. The block is a square with sides of length 1

Mashcka [7]

So if 11 units on the map is 5m in real life, the scale is 11 units= 5m
But we need 100m.
We would need 20 groups of 5m to get 100m.
So we would need 20 groups of 11 units, which is 220 units.
So, on the scale drawing, the square would need to be 220 units by 220 units.

5 0

3 years ago

Read 2 more answers

A view that shows half of an object as a sectional view and the other half of the object as a normal view is called a __________

Troyanec [42]

I think, it is half section.

6 0

3 years ago

Mariah collects 3 spiders each spider has eight legs.Brian gives her 1more spider. how many legs do Mariah's spiders have altoge

kompoz [17]

Tem 32 pernas de aranha

6 0

3 years ago

We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab

777dan777 [17]

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of $P(\bar X>0.50)$ is 0.50.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$