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3 years ago

(8-9)-(9+3)5 5+(5+5+6)+9 (5/3-3) 23+(4*2)/2

Mathematics

1 answer:

pashok25 [27]3 years ago

6 0

Answer:

If you need any further help, feel free to let me know

Step-by-step explanation:

The answers are

(8-9)-(9+3)*5=-61

5+(5+5+6)+9=30

(5/3-3)= -4 over 3 or -4/3 or -1.3 or -1 1/3 depends on what your choices are but all those are correct

2*3+(4*2)/2=10

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Step-by-step explanation:

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We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :

Null hypothesis: $\mu \leq 75$

Alternative hypothesis: $\mu > 75$

A. One-Tailed

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$

$p_v =P(z>2.189)=0.0143$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

We have the following data: 84 94 80 88 77 68 90 74 96 71

We can calculate the sample mean with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X=82.2$ represent the sample mean

$\sigma=10.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =75$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :

Null hypothesis: $\mu \leq 75$

Alternative hypothesis: $\mu > 75$

A. One-Tailed

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$

P-value

Since is a ONE-TAILED test the p value would given by:

$p_v =P(z>2.189)=0.0143$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

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3 years ago

(8-9)-(9+3)*5 5+(5+5+6)+9 (5/3-3) 2*3+(4*2)/2

(8-9)-(9+3)5 5+(5+5+6)+9 (5/3-3) 23+(4*2)/2