All categories

4 years ago

27,009,045 in expanded notation

Mathematics

2 answers:

rosijanka [135]4 years ago

8 0

20,000,000 + 7,000,000,+ 9,000 + 45
x

DaniilM [7]4 years ago

5 0

27,000,000 + 9,000 + 40 + 5

You might be interested in

Ms. Garcia watched 6 commercials

Aleonysh [2.5K]

Answer:

15 commercials

Step-by-step explanation:

6 commercials --> 24 minute

x commercials --> 60 minute(1 hour)

x*24 = 6*60

clear x

x = (6*60)/24

x= 15

6 0

3 years ago

What is the result when the number 79 is decreased by 17%?

trapecia [35]

Answer:

65.57 should be the correct answer!

Step-by-step explanation:

Here is the whole work I did, hope this helps!

100 - 17 = 83

79 x 0.83 = 65.57

8 0

3 years ago

Read 2 more answers

Help me plz i need it

maxonik [38]

Answer:

Its's the letter B 24cm²

6 0

3 years ago

Read 2 more answers

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

First we need to calculate the degrees of freedom given by:

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$

We can calculate also the p value:

$p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

5 0

3 years ago

Is this isosceles triangle acute or equiangular?

Artemon [7]

Answer:

Equiangular triangle

Step-by-step explanation:

3 0

3 years ago