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aleksklad [387]

2 years ago

10

The Dowers make pumpkin pie. If each person wants three slices of pie,and each pie is Cut into 9 slices, how many pies do they n

eed to make for 15 people?

1 answer:

Anestetic [448]2 years ago

5 0

Answer:

They need to make 5 pies for 15 people.

Step-by-step explanation:

15*3=45 slices of pie in total,

each pie is cut into 9 slices,

45/9=5 pies

You might be interested in

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

a pool is in the shape of a rectangular prism holds 1800 cubic feet of water. the pool is 6 feet deep and 10 feet wide. what is

nadya68 [22]

Answer:

Step-by-step explanation:

length = 1800 / 6/10 = 30 ft long = 10 yard

7 0

2 years ago

World wind energy generating1 capacity, W , was 371 gigawatts by the end of 2014 and has been increasing at a continuous rate of

Sunny_sXe [5.5K]

Answer:

a) $W(t) = 371(1.168)^{t}$

b) Wind capacity will pass 600 gigawatts during the year 2018

Step-by-step explanation:

The world wind energy generating capacity can be modeled by the following function

$W(t) = W(0)(1+r)^{t}$

In which W(t) is the wind energy generating capacity in t years after 2014, W(0) is the capacity in 2014 and r is the growth rate, as a decimal.

371 gigawatts by the end of 2014 and has been increasing at a continuous rate of approximately 16.8%.

This means that

$W(0) = 371, r = 0.168$

(a) Give a formula for W , in gigawatts, as a function of time, t , in years since the end of 2014 . W= gigawatts

$W(t) = W(0)(1+r)^{t}$

$W(t) = 371(1+0.168)^{t}$

$W(t) = 371(1.168)^{t}$

(b) When is wind capacity predicted to pass 600 gigawatts? Wind capacity will pass 600 gigawatts during the year?

This is t years after the end of 2014, in which t found when W(t) = 600. So

$W(t) = 371(1.168)^{t}$

$600 = 371(1.168)^{t}$

$(1.168)^{t} = \frac{600}{371}$

$(1.168)^{t} = 1.61725$

We have that:

$\log{a^{t}} = t\log{a}$

So we apply log to both sides of the equality

$\log{(1.168)^{t}} = \log{1.61725}$

$t\log{1.168} = 0.2088$

$0.0674t = 0.2088$

$t = \frac{0.2088}{0.0674}$

$t = 3.1$

It will happen 3.1 years after the end of 2014, so during the year of 2018.

7 0

2 years ago

Prepare the Sales Budget assuming: 1. Expected sales volume: 8,000 units for the first quarter, an increase of 20% is expected f

solong [7]

Answer:

Sales budget $ for the 4 quarters is $ 600,000 $ 720,000 $594000 and $ 825,000

Production Budget in units for the four quarters i s 7997.6 9000 8900 and 10,350 units

Step-by-step explanation:

Sales Budget

Quarters I II III IV

Sales Volume 8000 9600 7200 10,000

<u>Sales Price $75.00 $75.00 $ 82.50 $82.50 </u>

<u>Sales $ 600,000 720,000 594000 825,000 </u>

We calculate the sales volume by applying the given percent to the sales units for each quarter. Then multiply it with the sales price to get the sales budget.

<u />

Production Budget

Quarters I II III IV

Sales Volume 8000 9600 7200 10,000

+ Desired

Ending Inv. 2400 1800 3500 3850

<u>Less Opening ----- 2400 1800 3500 </u>

<u>Production 7997.6 9000 8900 10,350 </u>

We find the production budget units by adding the desired ending inventory and subtracting the opening inventory from the sales units calculated above. The ending inventory of one quarter is the opening inventory of the next quarter.As we do not know the ending inventory of the previous year we cannot find the opening inventory of the 1st quarter of the year .

5 0

2 years ago

Marie currently has a collection of 58 stamps if she buys s stamps each week for w Weeks which expression represents the total n

laila [671]

58 stamps is the number she gets each week. So if she went for one week, 58*1=58. 58*2=116 and so on. Because we don't know the exact number of weeks, we say 58w or 58*w because you multiply however many number of weeks she collects.

8 0

2 years ago