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Mariulka [41]
3 years ago
5

Please help!! I'll give 30 points to the brainliest!! Please help!!

Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0
Hey there :)

1st Shape:
A combination of a trapezium and a rectangle
We know the area formulas of rectangles → Area = length × width |
And Trapeziums → Area = \frac{1}{2} (base1+base2)(height)

We are given:
Length of rectangle and base 2 of trapezium = 20 in
Width of rectangle = 16 in
Base 1 of trapezium = 9 in
Height of trapezium = 21 - 16 = 5 in

Apply the formulas
Area of rectangle = 20 × 16 = 320 in²
Area of trapezium = \frac{1}{2} ( 9 + 20 )(5) = 72.5 in²

Add both areas if necessary = 320 + 72.5 = 392.5 in²

2nd Shape:
It is a trapezium

We know the area formula of a trapezium
Area = \frac{1}{2}(base 1 + base 2 )(height)

We know the
Base 1 = 8 units
Base 2 = 16 units
Height = 14 units

Apply the formula
Area = \frac{1}{2} (8+16)(14) = 168 units²
         
3rd Shape:
It is a trapezium

We know the:
Base 1 = 7 units
Base 2 = 24 units
Height = 24 units

Apply the formula
Area = \frac{1}{2}(7+24)(24) = 372 units²

4th Shape:
It is a trapezium

We know the:
Base 1 = 4 units
Base 2 = 8 units
Height = 6 units

Apply the formula:
Area = \frac{1}{2} ( 4 + 8 )(6) = 36 units²
   


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P-value=0.042.

Step-by-step explanation:

The question is incomplete:

The data of the scores for each student is:

Before    After

430        465

485        475

520        535

360        410

440        425

500        505

425        450

470        480

515        520

430        430

450        460

495        500

540        530

We will generate a sample for the difference of scores (before - after) and test that sample.

The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]

This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.

The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

Then, the null and alternative hypothesis are:

H_0: \mu=0\\\\H_a:\mu> 0

The significance level is 0.01.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{18.423}{\sqrt{13}}=5.11

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9.615-0}{5.11}=\dfrac{9.615}{5.11}=1.882

The degrees of freedom for this sample size are:

df=n-1=13-1=12

This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.882)=0.042

As the P-value (0.042) is bigger than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

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