Is the sum of two integers positive, negative, or zero?How can you tell?

Tim and Dan were playing football against Jason and Luke.Touchdowns are worth 7 points .Tim and Dad scored 4 touchdowns. Jason a

makkiz [27]

Answer:

35, if Tim and Dan scored 4 together, 7 if each scored 4 touchdowns

6 0

3 years ago

A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st

maxonik [38]

Answer:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

$p_v =2*P(z>1.413)=0.158$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample

$n=42$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t 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 true mean is different from 5, the system of hypothesis would be:

Null hypothesis: $\mu = 5$

Alternative hypothesis: $\mu \neq 5$

If we analyze the size for the sample is > 30 but and we know the population deviation so 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:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z>1.413)=0.158$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

8 0

3 years ago

Identify any x intersepts and y intersepts

Angelina_Jolie [31]

The x-intercept of the line is (-5,0) and the y-intercept is (0,5)

6 0

3 years ago

Answer ASAP PLS -

Talja [164]

TQV: 50
good luck :’)

3 0

3 years ago