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Crank
3 years ago
10

The value of a $60,000 tractor decreases approximately 20% each year. Find the approximate value of the tractor in 5 years.

Mathematics
2 answers:
Ad libitum [116K]3 years ago
8 0
The tractor will be worth 19660.8$ over the 5 years
777dan777 [17]3 years ago
7 0
Year 1: 60000*80% (to find value after 20% decrease. 100-20=80%)
60000*0.8 = $48000

Yr 2: 48000*0.8 = $38400
Yr3: 38400*0.8 = $30720
Yr 4: 30720*0.8 = $24576
Yr 5: 24576*0.8 = $19660.80 (value of tractor after 5 yrs of deprecation)
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Answer:

The range of crying times within 68% of the data  is (5.9, 8.1).

The range of crying times within 95% of the data  is (4.8, 9.2).

The range of crying times within 99.7% of the data  is (3.7, 10.3).

Step-by-step explanation:

According to the Empirical Rule in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:

  • 68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤  µ + σ) = 0.68.
  • 95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤  µ + 2σ) = 0.95.
  • 99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

The mean and standard deviation are:

µ = 7

σ = 1.1

Compute the  range of crying times within 68% of the data as follows:

P(\mu-\sigma\leq X\leq \mu+\sigma)=0.68\\\\P(7-1.1\leq X\leq 7+1.1)=0.68\\\\P(5.9\leq X\leq 8.1)=0.68

The range of crying times within 68% of the data  is (5.9, 8.1).

Compute the  range of crying times within 95% of the data as follows:

P(\mu-2\sigma\leq X\leq \mu+2\sigma)=0.95\\\\P(7-2.2\leq X\leq 7+2.2)=0.95\\\\P(4.8\leq X\leq 9.2)=0.95

The range of crying times within 95% of the data  is (4.8, 9.2).

Compute the  range of crying times within 99.7% of the data as follows:

P(\mu-3\sigma\leq X\leq \mu+3\sigma)=0.997\\\\P(7-3.3\leq X\leq 7+3.3)=0.997\\\\P(3.7\leq X\leq 10.3)=0.997

The range of crying times within 99.7% of the data  is (3.7, 10.3).

8 0
3 years ago
3) The population of a type of bacteria, starts with 100, and grows at a rate of 300% every
yawa3891 [41]
A) 100 • 3 • x
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3 years ago
The length of a rectangle is 2cm, and the width is 1 cm. Find the length of a diagonal of this rectangle.
pshichka [43]

Answer:

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Step by step explanation :

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7 0
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(7 4/9 −8)∙3.6–1.6∙( 1/8 − 3/4 )+1 2/5 ÷(−0.35)
lidiya [134]

Answer:

-7

Step-by-step explanation:

(7 4/9 −8)∙3.6–1.6∙( 1/8 − 3/4 )+1 2/5 ÷(−0.35)

PEMDAS, parenthesis

-5/9∙3.6–1.6∙(1/8-6/8)+1 2/5 ÷(−0.35)

PEMDAS, division

-5/9∙3.6–1.6∙-5/8-4

PEMDAS, multiplication

-5/9∙ 18/5– 8/5 ∙-5/8-4

-2-1-4

PEMDAS, subtraction

-7

Hope this helps! Please mark brainliest :)

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4 years ago
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