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3 years ago

Please help me with this if u get it right u get BRAINLISTED HELP PLS

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

Step-by-step explanation:

simplify:

3x+9>x+33

2x>24

x>12

now see the graphs, and B and D both are marked with 12

but D has no filled circle, so that is the right answer

I hope this helped, and if my answer was right, please mark this as brainliest! Thank you!

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The healthcare provider prescribes an IV solution of ampicillin 1 gram in 50 mL of DşW at 25 drops/minute. The nurse uses an inf

Morgarella [4.7K]

The infusion will be completed at 19:31

The medicine that the nurse is giving the patient is 1 gram of ampicillin diluted in 50 mL of dextrose water.

This means 50 mL + 1 mL = 51 mL. 1 gram of ampicillin = 1 mL considering all water-based substances have a density of 1gm/mL.

Now, the set delivers 15 ggt or drops per mililiter. Let's calculate the total drops in the 51 mL.

51mL x 15ggt =765 drops

Finally, we know the medicine is given at a rate of 25 drops per minute. Let's calculate the time:

765 drops / 25 drops per minute = 30.6 minutes

This means the medicine it will take 30.6 minutes for the infusion to be completed and therefore the infusion will be completed at 19:31 (19:00 + 30.6 minutes = 19:30.6 or 19:31).

Learn more in: brainly.com/question/25281327

5 0

2 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of530 bags per hour.(a-1) At α = .05 in

-Dominant- [34]

Answer:

b. H0: μ < 530. Reject H0 if tcalc > -1.753

Step-by-step explanation:

1) Data given and notation

$\bar X=507$ represent the sample mean

$s=47$ represent the sample standard deviation

$n=16$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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:

$t=\frac{507-530}{\frac{47}{\sqrt{16}}}=-1.957$

Critical value

Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:

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

Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be $t_{\alpha}=-1/753$ .

And we can use the following excel code to find it: "=T.INV(0.05,15)"

So then the correct rejection zone for H0 would be: Reject H0 if $t_{calc}$

P-value

Since is a one side left tailed test the p value would be:

$p_v =P(t_{(15)}$

Conclusion

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.