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olga55 [171]
3 years ago
13

R-(-4x - (y + y)); use x = -4, and y = -3 evaluate ​

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

insert values of x and y

r - ( -4(-4) - ( -3-3))

r - ( 16 +6 )

r - 22

Doneeeeeeeeeeeeeeeeee3

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The answer would be 16.
 

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Original: $300; new: $200
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You lost $100 because 300-200 is 100.
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5x - x = -12 what is x???
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X would equal -3. 5x-3 would equal -15, and 2 negatives would equal a positive. So, -15+3 would equal -12.
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The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
uppose we want to build a rectangular storage container with open top whose volume is $$ cubic meters. Assume that the cost of m
Ainat [17]

Answer:

a = length of the base = 2.172 m

b = width of the base = 1.357 m

c = height = 4.072 m

Step-by-step explanation:

Suppose we want to build a rectangular storage container with open top whose volume is 12 cubic meters. Assume that the cost of materials for the base is 12 dollars per square meter, and the cost of materials for the sides is 8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?

lets call a = length of the base

b = width of the base

c = height

V = a.b.c = 12

Area without the top:

Area = ab + 2bc + 2ac

Cost  = 12ab + 8.2bc + 8.2ac

Cost = 12ab + 16bc + 16ac

height = 3.width

c = 3b

Cost = 12ab + 16b.3b + 16a.3b = 12ab + 48b² + 48ab = 48b² + 60ab

abc = 12 → ab.3b = 12 → 3ab² = 12 → ab² = 4 → a = 4/b²

Cost = 48b² + 60ab = 48b² + 60b.4/b² = 48b² + 240/b

C(b) = 48b² + 240/b

C'(b) = 96b - 240/b²

Minimum cost: C'(b) = 0

96b - 240/b² = 0

(96b³ - 240)/b² = 0

96b³ - 240 = 0

96b³ = 240

b³ = 240/96

b³ = 2.5

b = 1.357m

c = 3b = 3*1.357 = 4.072m

a = 4/b² = 2.172m

6 0
3 years ago
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