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insens350 [35]
1 year ago
6

At a certain company, the annual winter party is always held on the second of Friday of December. What is the latest possible da

te for the party
Mathematics
1 answer:
vitfil [10]1 year ago
7 0

Answer:

The latest possible date is December, 14th.

Step-by-step explanation:

Notice that the second Friday of December is n+7, where n is the date for the first Friday of December. So, the latest the first Friday, the latest the second. As the weeks have seven days, the first Friday will be between 1st and 7th of each month. So, the latest first Friday will be 7th. Therefore, the latest second Friday will be 14th.

You might be interested in
It costs 20 cents to receive a photo and 30 cents to send a photo from a cellphone. C is the cost of one photo (either sent or r
ddd [48]

Answer:

(a) PC(C)=     \left \{ {{0.6 \ \ \ \ x=20} \atop 0.4 \ \ \ \ {x=30}} \right. \\\ 0 \ \ \ \ \ \ \ else

(b) E[C] = 24 cents

Step-by-step explanation:

Given:

Cost to receive a photo = 20 cents

Cost to send a photo = 30 cents

Probability of receiving a photo = 0.6

Probability of sending a photo = 0.4

We need to find

(a) PC(c)

(b) E[C]

Solution:

(a)

PC(C)=     \left \{ {{0.6 \ \ \ \ c=20} \atop 0.4 \ \ \ \ {c=30}} \right. \\\ 0 \ \ \ \ \ \ \ else

(b)

Expected value can be calculated by multiplying probability with cost.

E[C] = Probability × cost

E[C] = 0.6\times20 +0.4 \times 30 = 12 + 12 = 24\  cents

3 0
1 year ago
Last week, you spoke with 800 customers in 40 hours."
Masteriza [31]
The answer would be 20 im pretty sure
7 0
1 year ago
Two galaxies are shown. Galaxy U: Dust, gas, and stars in the shape of a pinwheel, relatively small in size. Galaxy NGC 1427A: D
Pani-rosa [81]

Answer:

NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in that both contain plenty of dust and gas. Both also have active star-forming sites.

Step-by-step explanation: Got it right on the test

8 0
2 years ago
Read 2 more answers
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equ
kompoz [17]

Answer:

0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution.

Problems of normally distributed samples can be 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.

In this problem, we have that:

\mu = 800, \sigma = 40, n = 16, s = \frac{40}{\sqrt{16}} = 10

Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

This probability is the pvalue of Z when X = 775. So

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

Z = \frac{775 - 800}{10}

Z = -2.5

Z = -2.5 has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.

5 0
2 years ago
Based on the information below, choose the correct answer.
Arte-miy333 [17]
The dollar amount of each monthly payment is interest
7,865.87÷120=65.55

The percent of the total payments is total interest
(7,865.87÷27,865.87)×100=28.2%
5 0
1 year ago
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