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3 years ago

Someone please help me with this geometry question

Mathematics

2 answers:

shutvik [7]3 years ago

7 0

Answer:

A parallelogram

Step-by-step explanation:

The opposite sides and angles are congruent.

Makovka662 [10]3 years ago

7 0

It’s a rhombus.

It’s a special parallelogram since all the sides are also equal in length

You might be interested in

A manufacturer of running shoes knows that the average lifetime for a particular model of shoes is 15 months. Someone in the res

AveGali [126]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the designer's claim of a better shoe is supported by the trial results.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 15 \ months$

The sample size is n = 25

The sample mean is $\= x = 17 \ months$

The standard deviation is $s = 5.5 \ months$

Let assume the level of significance of this test is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 15$

The alternative hypothesis is $H_a : \mu > 17$

Generally the degree of freedom is mathematically represented as

$df = n -1$

=> $df = 25 -1$

=> $df = 24$

Generally the test statistics is mathematically represented as

$t = \frac{\= x- \mu }{\frac{s}{\sqrt{n} } }$

=> $t = \frac{ 17 - 15 }{\frac{5.5}{\sqrt{25} } }$

=> $t = 1.8182$

Generally from the student t distribution table the probability of obtaining $t = 1.8182$ to the right of the curve at a degree of freedom of $df = 24$ is

$p-value = P(t > 0.18182 ) = 0.4286$

From the value obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the designer's claim of a better shoe is supported by the trial results.

7 0

2 years ago

Use partial fraction expansion to evaluate: LaTeX: \int\frac{x-1}{x^2+3x+2}dx ∫ x − 1 x 2 + 3 x + 2 d x a. LaTeX: -2\ln\left|x+1

ohaa [14]

$\frac{x-1}{x^2+3x+2}\\\frac{x-1}{(x+1)(x+2)}\\\frac{x-1}{(x+1)(x+2)}=\frac{A}{x+2}+\frac{B}{x+1}\\x-1=A(x+1)+B(x+2)\\\\\text{ Let } x=-1 \text{ then } -2=A(0)+B(-1+2) \text{ which means } B=-2\\\\\text{ Instead of choose } x \text{ to be -1, let's now set it equal to -2 to find } A\\\\-3=A(-1)+B(-2+2)\\-3=-A\\A=3\\\\ \text{ So your integral becomes } \\\int \frac{3}{x+2} dx+\int \frac{-2}{x+1} dx\\3 \ln|x+2|-2\ln|x+1|+C\\\\or\\-2\ln|x+1|+3\ln|x+2|+C$