All categories

3 years ago

2/9 long division showed

Mathematics

2 answers:

Alexxandr [17]3 years ago

7 0

$\dfrac{2}{9}=2\div9=0.222...=0.\overline{2}$

Remember:

$\dfrac{1}{9}=0.\overline{1}\\\\\dfrac{2}{9}=0.\overline{2}\\\\\dfrac{3}{9}=0.\overline{3}\\\\\dfrac{4}{9}=0.\overline{4}\\\\\dfrac{5}{9}=0.\overline{5}\\\\\dfrac{6}{9}=0.\overline{6}\\\\\dfrac{7}{9}=0.\overline{7}\\\\\dfrac{8}{9}=0.\overline{8}$

SIZIF [17.4K]3 years ago

3 0

recall that a/b to decimal is just a ÷ b, so 2/9 is just 2 ÷ 9.

check the picture below.

You might be interested in

Find the missing side. Round to the nearest tenth

Basile [38]

Answer:

A,c,b

Step-by-step explanation:

8 0

2 years ago

Find the coordinates of the intersection of diagonal of the parallelogram

IceJOKER [234]

Answer:

your mom

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

-(4t^2-5t-6) How to factor it?

swat32

How to factor it?
-(4t^2-5t-6)
=
-(t-2)(4t+3)

Try Mathway .com

8 0

3 years ago

Apply the distributive property to factor out the greatest common factor. 16a + 48

densk [106]

The distributive property is:
x*y+x*z=x*(y+z)

16a+48=
16a+16*3=
16*(a+3)

5 0

3 years ago

DDT is a pesticide banned in the United States for its danger to humans and animals. In an experiment on the impact of DDT, six

UNO [17]

Answer:

The p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to don't reject the claim that the group with DDT have a mean greater than the group for rats not poisoned.

Step-by-step explanation:

1) Data given and notation

P:[12.207 ,16.869, 25.050, 22.429, 8.456, 20.589]

UP:[11.074, 9.686 ,12.064, 9.351, 8.182, 6.642]

$\bar X_{P}=17.6$ represent the mean for the sample poisoned

$\bar X_{UP}=9.50$ represent the mean for the sample unpoisoned

$s_{P}=6.34$ represent the sample standard deviation for the sample poisoned

$s_{UP}=1.95$ represent the sample standard deviation for the sample unpoisoned

$n_{P}=6$ sample size for the group poisoned

$n_{UP}=6$ sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for rats exposed to DDT is greater than that for rats not poisoned , the system of hypothesis would be:

Null hypothesis: $\mu_{P} \leq \mu_{UP}$

Alternative hypothesis: $\mu_{P} > \mu_{UP}$

If we analyze the size for the samples both are less than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{P}-\bar X_{UP}}{\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{17.6-9.5}{\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}}=2.99$

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{P}+n_{UP}-2=6+6-2=10$

Since is a unilateral test the p value would be:

$p_v =P(t_{(10)}>2.99)=0.0067$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to don't reject the claim that the group with DDT have a mean greater than the group for rats not poisoned.

8 0

3 years ago