Subtract. (6x + 5) - (x + 3) A. 5x + 2 B. 6x + 2 C! 6x + 8 D. 5x + 8

Mathematics

2 answers:

Len [333]2 years ago

5 0

Answer:

Its A. 5x + 2

Step-by-step explanation:

a p e x ! <3

Natali [406]2 years ago

4 0

Answer: 5x + 2

Step-by-step explanation:

You might be interested in

For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test

Basile [38]

Answer:

a) $p_v =P(Z>2.05)=1-P(z$

b) $p_v =P(Z>-1.84)=1-P(z$

c) $p_v =P(Z>0.4)=1-P(z$

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: $\mu=10$

Alternative hypothesis: $\mu >10$

If the random variable is distributed like this: $X \sim N(\mu,\sigma)$

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

$z_o=\frac{\bar X -\mu}{\sigma}$

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

$p_v =P(Z>2.05)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

$p_v =P(Z>-1.84)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

$p_v =P(Z>0.4)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say $\alpha=0.05$ . For part a the $p_v$ so then we can reject the null hypothesis at this significance level.