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2 years ago

15

The Revenue for the week is $2000, and labor cost consists of two workers earning $8 per hour who work 40 hours each, what is la

bor cost as a percentage of Revenue?

1 answer:

kkurt [141]2 years ago

8 0

8*40 = 320 * 2 = $640 for the 2 workers

640/2000 = 0.32

labor cost is 32% of the revenue

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Harrison Water Sports has two retail outlets: Seattle and Portland. The Seattle store does 60 percent of the total sales in a ye

Anna11 [10]

Answer: P = 0.75

Step-by-step explanation:

Hi!

The sample space of this problems is the set of all the possible sales. It is divided in the disjoint sets:

$S_s = {\text{sales made in Seattle }}\\S_p={\text {sales made in Portland}}$

We have also the set of sales of boat accesories $S_b$ , the colored one in the image.

We are given the data:

$P(S_s) = 0.6\\P(S_b | S_s) = \frac{P(S_b\bigcap S_s)}{P(S_s)}=0.4\\P(S_b|S_p) =\frac{P(S_b\bigcap S_p)}{P(S_p)}=0.2$

From these relations you can compute the probabilities of the intersections colored in the image:

$pink\;set:\;P(S_b \bigcap S_s) =0.6*0.4=0.24\\blue\;set\;:P(S_b \bigcap S_p)=(1-0.6)*0.2 =0.08$

You are asked about the conditional probability:

$P(S_s|S_b) = \frac{P(S_s \bigcap S_b)}{P(S_b)}$

To calculate this, you need $P(S_b)$ . In the image you can see that the set $S_b$ is the union of the two disjoint pink and blue sets. Then:

$P(S_b)=P((S_b \bigcap S_s)\bigcup(S_b \bigcap S_p)) = 0.24 + 0.08 = 0.32$

Finally:

$P(S_s|S_b) = \frac{0.24}{0.32}=0.75$

4 0

2 years ago

A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age chil

Rom4ik [11]

Answer:

Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

$P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......$

where, n = number of trials (samples) taken = 20 children

r = number of success = none of the 20 children

p = probability of success which in our case is probability that

children are unvaccinated, i.e; p = 13.5%

<u><em>Let X = Number of children that are unvaccinated</em></u>

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

P(X = 0) = $\binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}$

= $1\times 1 \times 0.865^{20}$

= 0.055

<em>Hence, the probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.</em>

8 0

2 years ago

A prescription calls for the following: Sodium citrate 5 g Tartar Emetic 0.015 g Cherry syrup ad 120 mL Using a balance with a s

ella [17]

Answer:

measure the volume of cherry syrup solution with Tartar Emetic = 22.5 mL

Hence, the correct quantity of tartar emetic to fill the prescription will be obtained

Step-by-step explanation:

Given:

Mass of Sodium citrate = 5 g

Mass of Tartar Emetic = 0.015 g = 15 mg

Volume of Cherry syrup ad 120 mL = 120 mL

sensitivity = 4 mg

Error = 5%

Now,

The smallest weight that can be measured = $\frac{\textup{Sensitivity}}{\textup{Error}}\times100$

or

The smallest weight that can be measured = $\frac{\textup{4}}{\textup{5}}\times100$ = 80 mg

Now,

80 mg is dissolved in 120 mL of Cherry syrup according to prescription

( Thus,

1 mg is dissolved in $\frac{120}{80}\ mL$ of Cherry syrup according to prescription )

so to measure 15 mg of Tartar Emetic, measure the volume of cherry syrup

= $15\times\frac{120}{80}\ mL$

= 22.5 mL

Hence, the correct quantity of tartar emetic to fill the prescription will be obtained

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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ANEK [815]

Answer:

Step-by-step explanation:

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2 years ago