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3 years ago

I need help with this math

Mathematics

2 answers:

Tems11 [23]3 years ago

8 0

15/15 or 1 is your answer.

Georgia [21]3 years ago

8 0

15/15 or 1 is your answer

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X/2-x=x/4+6<br> I need help but it’s not ASAP

Ivanshal [37]

Answer is -8
(photo shows step by step)

7 0

3 years ago

No more asking me to click on links answer truthfully

Alborosie

Ok so the correct answers are both A. and B.

It would be A. between 1.5 and 2, because the number 1.77 is in between those two numbers. It is greater than 1.5 but also less than 2.

It would be B. to the right of 1.71, because the number 1.77 is greater than 1.71.

I hope this helps! Good luck on your homework!!

4 0

3 years ago

HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x

Anna35 [415]

$\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0$

$\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}$

3 0

4 years ago

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

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