Monica deposits $100 into a savings account that pays a simple interest rate of 4.5%, Paul

Will mark brainliest!

ioda

Answer:

the answer is A

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Eliza falls asleep at 9:35 pm and wakes up 8 hours and 45 miunets later, what time does she wake uo

alexdok [17]

Answer: Eliza wakes up at 6:25 A.M.

Step-by-step explanation:

What I did is first used the 45 minutes to make the time 10:00 P.M. which leaves you with 8 hours and 25 minutes, then i added the 8 hours onto 10:00 P.M. which leaves you at 6:00 A.M. with 25 minutes left, then you add the 25 minutes onto 6:00 A.M. which puts it at 6:25 A.M.

6 0

3 years ago

What is 700 times 18

ad-work [718]

Answer:

12,600

Step-by-step explanation:

8 0

4 years ago

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If 3(r+300)=6, then what would be the value of r+300-2?

saveliy_v [14]

Solve for r in the first equation:

3(r+300) = 6

Use the distributive property:

3r + 900 = 6

Subtract 900 from both sides:

3r = -894

Divide both sides by 3:

r = -894 / 3

r = -298

Now you have r, replace r in the second equation and solve:

r +300 -2 =

-298 + 300 - 2 = 0

The answer is 0.

6 0

4 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

fenix001 [56]

Answer:

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ represent the value that we want to test

$\alpha$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".

3) Calculate the statistic

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

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$

4) Calculate the critical value

We need to begin calculating the degrees of freedom

$df=n-1=16-1=15$

The critical value for this case would be :

$P(t_{15}$

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}$

6) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

8 0

3 years ago