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4 years ago

Answer................::.

Mathematics

1 answer:

SashulF [63]4 years ago

7 0

<h3>Answer: 33</h3>

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Work Shown:

8^(x+7) = 32^(x-9)

(8)^(x+7) = (32)^(x-9)

(2^3)^(x+7) = (32)^(x-9) .... replace 8 with 2^3

(2^3)^(x+7) = (2^5)^(x-9) .... replace 32 with 2^5

2^(3(x+7)) = 2^(5(x-9)) ... multiply exponents

3(x+7) = 5(x-9) ...bases are the same, exponents are the same

3x+21 = 5x-45

21+45 = 5x-3x

66 = 2x

2x = 66

x = 66/2

x = 33

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The operations manager of a mail order house purchases double (D) and twin (T) beds for resale. Each double bed costs $500 and r

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Answer:

The weekly profit when ordering the optimal amounts is $42,000

Step-by-step explanation:

Cost of one double bed = $500, storage space of one double bed = 100 cubic feet

Cost of one twin bed = $300, storage space of one twin bed = 90 cubic feet

Capital for the week = $75,000

Storage capacity of warehouse = 18,000 cubic feet

Profit for each double bed = $300

Profit for each twin bed = $150

Since the capital for the week is $75,000 and the storage capacity of the warehouse is 18,000 cubic feet, the optimal amounts the operations manager can order is 90 double bed and 100 twin bed

Cost of ordering 90 double bed = 90 × $500 = $45,000

Storage space = 90 × 100 cubic feet = 9000 cubic feet

Cost of ordering 100 twin bed = 100 × $300 = $30,000

Storage space = 100 × 90 cubic feet = 9,000 cubic feet

Total cost = $45,000 + $30,000 = $75,000

Total storage space = 9,000 cubic feet + 9,000 cubic feet = 18,000

Therefore, weekly profit = (90 ×$300) + (100 × $150) = ($27,000 + $15,000) = $42,000

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3 years ago

5. The ratio of the number of John's stamps to the number of Peter's stamps is 5:8. Peter has 18 more stamps than John. If Peter

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Answer:

2:1

Step-by-step explanation:

J:P = 5:8

30:48

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2 years ago

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.

babunello [35]

Answer:

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

$\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535$

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002$

$P(X>0.8536)=P(z>0.002)=0.4992$

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288$

$P(X_s>0.8543)=P(z>0.3288)=0.37115$

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89$

$P(X_s>0.8556)=P(z>0.89)=0.18673$

The probability is P=0.187, so the claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

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3 years ago

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3 years ago

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