Find the product. write the in simplest form 1 1/7×1 3/4=

Mathematics

2 answers:

jolli1 [7]4 years ago

6 0

8. 7. 56.
- x. - = -- = 2
7. 4. 28

Ede4ka [16]4 years ago

6 0

$1\frac17\cdot\frac34=\frac{1\cdot7+1}{7}\cdot\frac34=\frac{8}{7}\cdot\frac34=\frac{24}{28}=\frac{6\cdot4}{7\cdot4}=\frac67$

You might be interested in

Test the claim below about the mean of the differences for a population of paired data at the level of significance α Assume the

Evgen [1.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is F

$t = 1.17$

$t_ {\alpha , df} =t_ {0.10 , 8} = 1.86$

Since the test statistics is outside the rejection region , we fail to reject the null hypothesis ,There is no statistically significant evidence to reject the claim

Step-by-step explanation:

From the question we are told that

The claim is $\mu = 0$

Hence

The null hypothesis is $H_o : \mu = 0$

The alternative is $H_a : \mu \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= d - \mu_d }{ \frac{s_d}{ \sqrt{n} } }$

=> $t = \frac{3.5 -0 }{ \frac{8.94}{ \sqrt{9} } }$

=> $t = 1.17$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

$df = 9 - 1$

$df = 8$

From the student t-distribution table the critical value of $\alpha$ at a degree of freedom of 8 is

$t_ {\alpha , df} =t_ {0.10 , 8} = 1.86$

Since the $t_ {\alpha , df}$ is outside the rejection region , we fail to reject the null hypothesis ,There is no sufficient evidence to reject the claim