What is 400 in expanded form (Warfare you get brainliest)

2 answers:

5 0

Hey there, Lets solve your question together

Question - What is 400 in Expanded form ?

We know that expanded form is a spread out way to write a number by showing the value of each individual digit that contributes to make the full number.

Lets use an example to show expanded form

Number - 4,265

= 4 x 1,000 + 2 x 100 + 6 x 10 + 5 x 1

Now lets solve for the number 400 in expanded form.

The number "4" is in the hundreds place

The number "0" is in the tenths place

The number "0" is in the One's place

Therefore 400 in expanded form is 400+00+0.

 Hope this helps

3241004551 [841]2 years ago

3 0

Hey There,

The answer is 400+00+0

Solution-
1) 4 is in the hundreds place (400)

2) 0 is in the tens place (00)

3) 0 is in the ones place (0)

Thus, the answer is 400+00+0

Hope this helps!

You might be interested in

A particular battery claims to have a mean life of 400 hours with a standard deviation of 30 hours. Approximately what percent o

Svetach [21]

Using the normal distribution, it is found that 25.14% of the batteries will last more than 420 hours.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, we have that the mean and the standard deviation are given, respectively, by:

$\mu = 400, \sigma = 30$ .

The proportion of the batteries will last more than 420 hours is one subtracted by the p-value of Z when X = 420, hence:

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

$Z = \frac{420 - 400}{30}$

Z = 0.67

Z = 0.67 has a p-value of 0.7486.

1 - 0.7486 = 0.2514.

0.2514 = 25.14% of the batteries will last more than 420 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

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3 0

1 year ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by