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3 years ago

This is just algebra need some help 10 pts each, will give brainliest

Mathematics

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

$\huge \boxed{ \boxed{z = 3}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

equation
PEMDAS

<h3>let's solve:</h3>

$\sf simplify \: substraction : \\ \frac{3}{ \frac{z - 2}{z} } = 3z$
$\sf simplify \: complex \: fractio n : \\ \frac{3z}{z - 2} = 3z$
$\sf cross \: multiplication : \\ 3z = 3z(z - 2)$
$\sf divide \: both \: sides \: by \: 3z : \\ \frac{3z}{3z} = \frac{3z(z - 2)}{3z} \\ 1 = z - 2$
$\sf add \: 2 \: to \: bot h \: sides : \\ z - 2 + 2 = 1 + 2 \\ \therefore \: z = 3$

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Answer:

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Read 2 more answers

What is the domain of the function shown on the graph?

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Step-by-step explanation:

The last one.

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3 years ago

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.

Kobotan [32]

Answer:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=3.7$ represent the mean for athletes

$\bar X_{NA}=4.1$ represent the mean for non athletes

$s_{A}=1.1$ represent the sample standard deviation for athletes

$s_{NA}=1.3$ represent the sample standard deviation for non athletes

$n_{A}=32$ sample size for the group 2

$n_{NA}=45$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{NA}\geq 0$

Alternative hypothesis: $\mu_{A} - \mu_{NA}< 0$

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

P value

We need to find first the degrees of freedom given by:

$df=n_A +n_{NA}-2=32+45-2=75$

Since is a one left tailed test the p value would be:

$p_v =P(t_{75}$

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see attachment for graph

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