All categories

3 years ago

Write 9983 in expanded form

Mathematics

2 answers:

stepladder [879]3 years ago

5 0

Answer:

Step-by-step explanation:

9,000 + 900 + 83 = 9.983

Over [174]3 years ago

4 0

Answer:

9,000 + 900 + 83

You might be interested in

jacobs puzzle has 84 pieces. Jacobs puts together 27 pieces in the morning. he puts together 38 more pieces in the afternoon. ho

elixir [45]

27 + 38 = 65

84 - 65 = 19

7 0

2 years ago

Read 2 more answers

Sam is a school leader. She wants to decide whether makeup should be allowed in school or not. She

anastassius [24]

Answer:

Following are the sample list can be attached as follows:

Step-by-step explanation:

In the given question some information is missing that is attachment of list which can be attached as follows:

please find the attachment list:

by evaluating the list it will give the answer that the both the samples most want to wear mascara the least want to wear lipstick.

8 0

3 years ago

I need this solved ASAP please ratio 4:12 simplified.

saveliy_v [14]

I’m not 100% certain but i’m pretty sure 1:3 works

5 0

3 years ago

Read 2 more answers

Graph this equation y=7/2x-2

Ludmilka [50]

To find the slope and y intercept, use the y=mx+b formula where m is the slope and b is the y intercept.
y=mx+b

Pull the values of m and b using the y=mx+b formula.
m=7/2,b=−2 where m is the slope and b is the y-intercept

6 0

3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

2 years ago