(3x+36)^2 can anyone please answer i don’t understand how to do this

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs

MissTica

Answer:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

Since is a right tailed test the p value would be:

$p_v =P(t_{25}>1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

$s=8.43$ represent the sample deviation

$n=26$ sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ 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 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu > 52$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

Since is a right tailed test the p value would be:

$p_v =P(t_{25}>1.8025)=0.0418$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

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3 years ago

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If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express y

zaharov [31]

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are independent, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for each of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of desired outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is at least one 5 rolled. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.