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2 years ago

5

A nurse is preparing to administer dextrose 5% water (D5W) 250 ml IV to infuse over 2 hr. The nurse should set the IV pump to de

liver how many ml/hr?

1 answer:

Molodets [167]2 years ago

4 0

Answer:

the answer is 125

Step-by-step explanation:

I divided 250 and 2

125***ANSWER

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The mass of a stone is 5kg. The stone is twice as heavy as a book. The book is 5 times heavier than a ball. What is the mass of

Irina18 [472]

The answer is 1.01 I did it on paper

8 0

2 years ago

Type the correct answer in the box. Round your answer to the thousandth place.

Alexandra [31]

Answer:

The probability that carbon emissions from the factory are within the permissible level and the test predicts the opposite to be true is 19.338% (Rounding to the next thousandth place)

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Probability that carbon emissions from the company’s factory exceed the permissible level = 35% = 0.35

Accuracy of the test of emissions level = 85% = 0.85

2. The probability that carbon emissions from the factory are within the permissible level and the test predicts the opposite to be true is?

These two events, carbon emissions from the company’s factory and the accuracy of the test are independent events, therefore:

Probability that carbon emissions from the factory are within the permissible level = 1 - 0.35 = 0.65

Probability that the test predicts the opposite to be true = 0..35 * 0.85 = 0.2975 (The opposite is that the carbon emissions from the company exceed the permissible level)

Probability that carbon emissions from the factory are within the permissible level and the test predicts the opposite to be true is:

0.65 * 0.2975 = 0.193375

The probability that carbon emissions from the factory are within the permissible level and the test predicts the opposite to be true is 19.338% (Rounding to the next thousandth place

MARK BRAINLIEST PLS

5 0

2 years ago

CAN SOMEONE PLEASE HELP ME RIGHT NOW.

exis [7]

Answer:

m∠SRV = 48°

Step-by-step explanation:

In the parallelogram attached,

m∠TUV = 78°

m∠TVU = 54°

By applying the property of the angles of a triangle in ΔTVU,

m∠TUV + m∠TVU + m∠UTV = 180°

78° + 54° + m∠UTV = 180°

m∠UTV = 180° - 132°

= 48°

Sides RS and TU are the parallel sides of the parallelogram and diagonal TR is a transverse.

Therefore, ∠UTV ≅ ∠SRV [Alternate interior angles]

m∠UTV = m∠SRV = 48°

6 0

2 years ago

The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random s

lidiya [134]

Answer:

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67$

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{73500 - 74000}{416.67}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

X = 71000

$Z = \frac{X - \mu}{s}$

$Z = \frac{71000 - 74000}{416.67}$

$Z = -7.2$

$Z = -7.2$ has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

8 0

2 years ago

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a

laila [671]

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

3 0

2 years ago