Please help! & don’t answer with “don’t know the answer”

What’s the answer? Because am having a hard time with this math problem.

Studentka2010 [4]

9514 1404 393

Answer:

34.5 square meters

Step-by-step explanation:

We assume you want to find the area of the shaded region. (The actual question is not visible here.)

The area of the triangle (including the rectangle) is given by the formula ...

A = 1/2bh

The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...

A = (1/2)(11 m)(9 m) = 49.5 m^2

The rectangle area is the product of its length and width:

A = LW

The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...

A = (5 m)(3 m) = 15 m^2

The shaded area is the difference between the triangle area and the rectangle area:

shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2

The shaded region has an area of 34.5 square meters.

7 0

2 years ago

I need help on numbers 17&18.

jekas [21]

I don't know if this is just my phone, but the picture won't load.

You could try posting a new picture? Or feel free to message me with the picture and I'll be glad to help! :)

5 0

2 years ago

Read 2 more answers

Please help! it’s due in 10 minutes i need help !!

icang [17]

Answer:

the answer is 4

Step-by-step explanation:

4 0

2 years ago

HELP ME PLEASE I NEED HELP

eduard

Answer:

sorry dont know

Step-by-step explanation:

6 0

2 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