Which of the following sequences of numbers are arithmetic sequences?

At barlow school 4/9 of the 873 students are boys at willow school 2/3 of the 630 students are girls which school has the greate

marishachu [46]

Answer:

There are 178 more boys in Barlow school as compared to Willow school.

Step-by-step explanation:

We are given the following in the question:

Barlow school:

Total number of students = 873

Number of boys =

$=\dfrac{4}{9}\times 873 = 388$

Willow school:

Total number of students = 630

Number of girls =

$=\dfrac{2}{3}\times 630 = 420$

Number of boys =

Total number of students - total number of girls

$=630-420 = 210$

Thus, Barlow school has greater number of boys than Willow school.

Difference between the number of boys =

$=388-210\\=178$

Thus, there are 178 more boys in Barlow school as compared to Willow school.

7 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

What is the slope of the line?

vampirchik [111]

Answer:

The slope is 2/1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What does 9/10-13/18 equal as an improper fraction<br> 100 points

Liono4ka [1.6K]

Answer:

$\huge \frac{8}{45}$

Step-by-step explanation:

$\frac{9}{10} - \frac{13}{18} \\ \\ = \frac{9 \times 9}{10 \times 9} - \frac{13 \times 5}{18 \times 5} \\ \\ = \frac{81}{90} - \frac{65}{90} \\ \\ = \frac{81 - 65}{90}\\ \\ = \frac{16}{90} \\ \\ = \frac{8}{45} \\$