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Oksanka [162]
3 years ago
11

Evaluate.....that.....idk how to write/say it lol

Mathematics
1 answer:
denis-greek [22]3 years ago
6 0

Answer:

4

Step-by-step explanation:

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
lord [1]

Given : A florist currently makes a profit of $20 on each of her celebration bouquets and sells an average of 30 bouquets every week . and graph

To Find :  Maximum profit , breakeven point , profit interval

Solution:

The maximum profit the florist will earn from selling celebration bouquets is $ 675

peak of y from Graph

The florist will break-even after  Selling 20     one-dollar decreases.

at breakeven

Break even is the point where the profit p(x) becomes 0

The interval of the number of one-dollar decreases for which the florist makes a profit from celebration bouquets is (0 ,20).

after 20 , P(x) is - ve

3 0
3 years ago
Read 2 more answers
URGENT PLEASE HELP<br> x/240 = 82/100 what is X?
Zielflug [23.3K]

x/240 = 82/100

240 : 100 = 2.4

so 240 is divided with 2,4 giving 100, this will work with 82 too. So, 82 × 2.4 = 196,8

Answer : x = 196,8

prove :

196,8/240 = 82/100 = 41/50

see the picture

7 0
3 years ago
Read 2 more answers
Helppppppppppppppppppppppppppppppppp!!!!!!!!!!!!!!
allsm [11]
∠DOC is the average of the given angle values,
  (170° +44°)/2 = 107°

∠DOB is the supplement to that,
  180° -107° = 73°

The appropriate choice is
  B)   73°
5 0
3 years ago
Keenan has 21 CDs and 33 DVDs. What is the ratio of DVDs to CDs?
mamaluj [8]
33: 21 ...........................................
7 0
3 years ago
Read 2 more answers
The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

Z = \frac{78 - 74}{2.5}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

Z = \frac{77 - 74}{1.2274}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

X = 71

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

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that n = 219, s = \frac{15}{\sqrt{219}} = 1.0136

This probability is the pvalue of Z when X = 74.2. So

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

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0
3 years ago
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