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3 years ago

5

Safety training in the workplace should be focused on _______.

2 answers:

Oksanka [162]3 years ago

4 0

Answer: Option 'a' is correct.

Step-by-step explanation:

Safety training in the workplace should be focused on " Prevention of accidents ".

Safety training is required or essential on that field where a small mistake can cause hazardous event.

Like Train operator, pilot etc.

A small mistake of them can kill lot of people and ruin their family.

Hence, Option 'a' is correct.

makvit [3.9K]3 years ago

3 0

The correct answer for the question that is being presented above is this one: "a. prevention of accidents." Safety training in the workplace should be focused on a. prevention of accidents. Prevention is better than cure. As early as possible, any untoward incidents must be prevented.

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3 years ago

How much does 60% out of 100

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Answer:

60

Step-by-step explanation:

Given: 60% out of 100

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3 years ago

A sphere has a radius of 7.5 centimeters. To the nearest cubic centimeter, what is the volume of the sphere?

Naya [18.7K]

Volume of sphere = 1766.25 cm^3

Step-by-step explanation:

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3 years ago

A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The v

BigorU [14]

Answer: B) [-0.0332,0.1332]

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval where we can be a percentage sure the true mean is.

The confidence interval for a difference in population proportion is calculated following these steps:

First, let's find population proportion for each population:

$p_{1}=\frac{30}{120}=0.25$

$p_{2}=\frac{32}{160}=0.2$

Second, calculate standard deviation for each proportion:

$\sigma_{1}=\sqrt{\frac{0.25(0.75)}{120} } = 0.0395$

$\sigma_{2}=\sqrt{\frac{0.2(0.8)}{160} } = 0.0316$

Now, we calculate standard error for difference:

$SE=\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}$

$SE=\sqrt{0.0395^{2}+0.0316^{2}}$

SE = 0.0505

The z-score for a 90% CI is 1.645.

Then, confidence interval is

$p_{1}-p_{2}$ ± z-score.SE

$0.25-0.2$ ± $1.645(0.0505)$

0.05 ± 0.0831

The limits of this interval are:

inferior: 0.05 - 0.0831 = $-0.0332$

superior: 0.05 + 0.0831 = 0.1332

The 90% confidence interval for the difference in the population proportion of pit pulls and golden retrievers is $[-0.0332,0.1332]$ .

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3 years ago

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