ABCDE is a pentagon.

statuscvo [17]

The data is a little bit complex,but you can use the knowledge of vector in the subject of algebra,or you can ultimate the basic Euclidean knowledge,like cosine theorem.In detail,you can list two equation with two variable,and get the answer.but i don't finish calculating it.

3 0

3 years ago

What is the slope intercept from of the equation of the line that that passes through the points (-3, 2) and (1, 5)

viva [34]

The slope-intercept form of a line:

$y=mx+b\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\to slope\\\\b\to y-intercept$

We have:

$(-3,\ 2)\to x_1=-3,\ y_1=2\\(1,\ 5)\to x_2=1,\ y_2=5$

Substitute:

$m=\dfrac{5-2}{1-(-3)}=\dfrac{3}{4}$

$y=\dfrac{3}{4}x+b$

Put the values of coordinates of the point (1, 5) to the equation of a line:

$5=\dfrac{3}{4}(1)+b\\\\5=\dfrac{3}{4}+b\qquad|-\dfrac{3}{4}\\\\b=4\dfrac{1}{4}$

Answer: $y=\dfrac{3}{4}x+4\dfrac{1}{4}$

5 0

3 years ago

A rectangular lot is 145 yards long and 100 yards wide.

mr_godi [17]

Answer:

150 by 100

Step-by-step explanation:

5 0

2 years ago

Help please

mrs_skeptik [129]

It will be 1.104 so i think it will be that ikd

6 0

3 years ago

Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c

Elina [12.6K]

Answer:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

Step-by-step explanation:

Data given

$n_1 =8$ represent the sample size for group Boston

$n_2 =9$ represent the sample size for group Dallas

$\bar X_1 =47$ represent the sample mean for the group Boston

$\bar X_2 =44$ represent the sample mean for the group Dallas

$s_1=3$ represent the sample standard deviation for group Boston

$s_2=3$ represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t student distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9$

And the deviation would be just the square root of the variance:

$S_p=3$

The statitsic would be:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

5 0

3 years ago