All categories

2 years ago

Calculate (a) the number of milligrams of metoclopramide HCl in each milliliter of the prescription:

Mathematics

1 answer:

nalin [4]2 years ago

7 0

Answer:

There are 100 milligrams of metoclopramide HCl in each milliliter of the prescription

Step-by-step explanation:

When the prescription says Purified water qs ad 100 mL means that if we were to make this, we should add the quantities given and then, fill it up with water until we have 100 mL of solution, being the key words qs ad, meaning sufficient quantity to get the amount of mixture given.

Then, knowing there is 10 grams of metoclopramide HCl per 100 mL of prescription, that means there is (1 gram = 1000 milligrams) 10000 milligrams of metoclopramide HCl per 100 mL of prescription. That is a concentration given in a mass/volume way.

Knowing the concentration, we can calculate it per mL instead of per 100 mL

$Concentration_{metoclopramide HCL}= \frac{10000mg}{100mL} =100 \frac{mg}{mL}$

You might be interested in

Please help, will mark brainliest

kogti [31]

Answer:

Step-by-step explanation:

⇒the slope is constant so the relationship will be linear

⇒the slope equal = (35-21)/(5-3)=7

(y-21)=7(x-3)

I hope this help

6 0

2 years ago

The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the e

Elodia [21]

Answer:

$F=\frac{s^2_2}{s^2_1}=\frac{0.8^2}{1.0^2}=0.64$

$p_v =P(F_{15,10}$

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the variation for the New sample it's significantly less than the variation for the Old sample at 5% of significance.

Step-by-step explanation:

Data given and notation

$n_1 = 11$ represent the sampe size for the Old

$n_2 =16$ represent the sample size for the New

$\bar X_1 =6.25$ represent the sample mean for Old

$\bar X_2 =5.95$ represent the sample mean for the New

$s_1 = 1.0$ represent the sample deviation for Old

$s_2 = 0.8$ represent the sample deviation for New

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for New sample it's lower than the variation for the Old sample, so the system of hypothesis are:

H0: $\sigma^2_2 \geq \sigma^2_1$

H1: $\sigma^2_2$

Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{0.8^2}{1.0^2}=0.64$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =11-1=10$ and the F statistic have 15 degrees of freedom for the numerator and 10 for the denominator. And the P value is given by: