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2 years ago

Solve this inequality.

Mathematics

1 answer:

Tomtit [17]2 years ago

7 0

x < −8/21

if you need help with explanations of the equations, "mathpapa" is something to look up. It's a algebra calculator that explains the step by step process

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A 1L IV contains 60meq of kcal. The IV was discontinued after 400ml has infused. How much lvl did the patient receive

Leya [2.2K]

If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.

Calculation for the Amount of IV

It is given that,

1L IV contains 60meq of kcal

⇒ 1000 mL of IV contains 60meq of kcal

⇒ 1 mL of IV will contain 60 / 1000 meq of kcal

As per the question,

The amount of IV infused in the patient = 400 mL

⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)

= 4 × 6

= 24 meq of kcal

Hence, the patient receives 24meq of kcal amount of IV.

Learn more about amount here:

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6 0

1 year ago

Write 4.51 as a mixed number and as an improper fraction.

anyanavicka [17]

it will be 451/100 as an improper fraction! :)

7 0

3 years ago

Use technology or a z-score table to answer the question.

Alik [6]

Answer:

The second choice: Approximately $65.2\%$ of the pretzel bags here will contain between 225 and 245 pretzels.

Step-by-step explanation:

This explanation uses a z-score table where each $z$ entry has two decimal places.

Let $\mu$ represent the mean of a normal distribution of variable $X$ . Let $\sigma$ be the standard deviation of the distribution. The z-score for the observation $x$ would be:

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question,

$\mu = 240$ .
$\sigma = 9.3$ .

Calculate the z-score for $x_1 = 225$ and $x_2 = 245$ . Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.

$\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}$ .

$\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}$ .

The question is asking for the probability $P(225 \le X \le 245)$ (where $X$ is between two values.) In this case, that's the same as $P(-1.61 \le Z \le 0.54)$ .

Keep in mind that the probabilities on many z-table correspond to probability of $P(Z \le z)$ (where $Z$ is no greater than one value.) Therefore, apply the identity $P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$ to rewrite $P(-1.61 \le Z \le 0.54)$ as the difference between two probabilities:

$P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)$ .

Look up the z-table for $P(Z \le 0.54)$ and $P(Z \le -1.61)$ :

$P(Z \le 0.54)\approx 0.70540$ .
$P(Z \le -1.61) \approx 0.05370$ .

$\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}$ .