Write the fraction 39 over 9th a repeating decimal

When we multiply a whole number with a fraction, we can think of the whole numbers as "over one"...

5 3 5 15
3 * --- = --- * ---- = ----
6 1 6 6

We can further simplify the fraction, by dividing numerator and denominator by 3...

15 15 / 3 5
--- = ------- = ----
6 6 / 3 2

Since this is an improper fraction, we need to turn it into a mixed number...

5 2 + 2 + 1 2 2 1 1 1
---- = -------------- = --- + ---- + --- = 1 + 1 + ---- = 2 --- or 2.5
2 2 2 2 2 2 2

6 0

3 years ago

What are the converse, inverse, and contrapositive of the following true conditional? What are the truth values of each?

ser-zykov [4K]

In logic, the converse of a conditional statement is the result of reversing its two parts. For example, the statement P → Q, has the converse of Q → P.

For the given statement, 'If a figure is a rectangle, then it is a parallelogram.' the converse is 'if a figure is a parallelogram, then it is rectangle.'
As can be seen, the converse statement is not true, hence the truth value of the converse statement is false.

The inverse of a conditional statement is the result of negating both the hypothesis and conclusion of the conditional statement. For example, the inverse of P → Q is ~P → ~Q.
For the given statement, 'If a figure is a rectangle, then it is a parallelogram.' the inverse is 'if a figure is not a rectangle, then it is not a parallelogram.'
As can be seen, the inverse statement is not true, hence the truth value of the inverse statement is false.

The contrapositive of a conditional statement is switching the hypothesis and conclusion of the conditional statement and negating both. For example, the contrapositive of P → Q is ~Q → ~P. 
For the given statement, 'If a figure is a rectangle, then it is a parallelogram.' the contrapositive is 'if a figure is not a parallelogram, then it is not a rectangle.'
As can be seen, the contrapositive statement is true, hence the truth value of the contrapositive statement is true.

4 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

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.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago