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Feliz [49]
3 years ago
14

The garden club

Mathematics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

here's a link

bit. ly/3gVQkw3

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Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

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

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

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

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

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

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
Solve using elimination 2x- 2y=-8 x+2y=-1
dimulka [17.4K]

Answer:( -3,1)

X=-3 y=1

Step-by-step explanation:

6 0
3 years ago
Which graph shows the solution to the inequality?<br><br> x - 4 ≤ -1
GenaCL600 [577]

Answer:

x -4 \leq  -1\\x -4 + 4 \leq -1 + 4\\x\leq -1 +4\\x\leq +3

4 0
2 years ago
Read 2 more answers
Suzanne Attired makes fashionable garments. During a particular week employees worked 360 hours to produce a batch of 132 garmen
lozanna [386]

Answer:

$57.44 per hour

Step-by-step explanation:

Given that :

Hours worked = 360 hours

Production = 132 batches of garment

Of the 152:

52 were flawed (sold at $90 each)

80 were sold at $200 each

Labor productivity ratio: (output value / input value (hours worked))

Output value = [(52*$90) + (80*$200)]

Output value = $20,680

Labor productivity ratio = $20,680 / 360 hours

Labor productivity ratio = $57.44 per hour

7 0
3 years ago
A sports agent made one million dollars by charging a 12.5% commission to negotiate a long-term contract for a professional athl
Hatshy [7]
$1,000,000 = 12.5% of the contract

"12.5%"  means  0.125

"of" means "times"

1,000,000 = 0.125 C

Divide each side by 0.125 :

C = 1,000,000 / 0.125 = <em>$8 million</em>




4 0
3 years ago
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