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2 years ago

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Government regulations indicate that the total weight of cargo in a certain kind of airplane cannot exceed 330 kg. On a particul

ar day a plane is loaded with 100 boxes of goods. If the weight distribution for individual boxes is normal with mean 3.2 kg and standard deviation 7 kg, what is the probability that the regulations will NOT be met:(a)- 1.5% (b)- 92% (c)- 8% (d)- 15% (e)- 85%

1 answer:

harkovskaia [24]2 years ago

6 0

Answer:

The probability that the regulations will NOT be met is (c)- 8%

Step-by-step explanation:

The standard deviation information in the question is wrong. It is not 7, It should be 0.7

the probability that the regulations will NOT be met for the plane loaded with 100 boxes of goods is

=P(X>3.3 kg) where 3.3 is the mean weight of the 100 boxes in the plane.

=P(z>z*) where z* is the statistic of 3.30 kg in the weight distribution.

z* can be calculated as

z*= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X is the cutoff mean weight of the boxes the airplane cannot exceed (3.3 kg)

M is the mean weight of the boxes (3.2)

s is the standard deviation (0.7 kg)

N is the sample size (100)

then z*= $\frac{3.3-3.2}{\frac{0,7}{\sqrt{100} } }$ ≈ 1.43

then P(z>z*)≈0.08 or 8%

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What property was used on 14k+2(3k+5)-5=10 to obtain 14k+6k+10-5=10

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The difference between

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This is called the DISTRIBUTIVE PROPERTY

Answer: DISTRIBUTIVE PROPERTY

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2 years ago

A greengrocer sells a melon at a profit of 37.5% on the price he pays for it. what is the ratio the cost price to the selling pr

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2 years ago

In a random sample of 8 people with advanced degrees in biology, the mean monthly income was $4744 and the standard deviation wa

eimsori [14]

Answer: u= ( 4342.08, 5145.92).

Step-by-step explanation: the population mean is estimated using the sample by the formulae assuming a 95% confidence level

u = x' + Zα/2 * (√σ/n) or x' - Zα/2 * (√σ/n)

u = estimated population mean

x' = sample mean = 4744

n = sample size =8

σ = sample standard deviation. = 580

α = level of significance = 1- confidence level = 1-0.95= 0.05

Zα/2 = z score from the normal distribution table for a 2 tailed test = 1.96

First boundary value for interval

u = 4744 + 1.96 ( 580/√8)

u = 4744 + 1.96 * (205.0609)

u = 4744 + 401.92

u = 5145.92

Second boundary value for interval

u = 4744 - 1.96 ( 580/√8)

u = 4744 - 1.96 * (205.0609)

u = 4744 - 401.92

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2 years ago

A treasure map says that a treasure is buried so that it partitions the distance between a rock and a tree in a 5:9 ratio. Marin

djyliett [7]

Answer:

A) (7.6, 8.8)

which are the coordinates of the treasure.

Missing Problem Statement:

Given Options:

A) (11.4, 14.2)

B) (7.6, 8.8)

C) (5.7, 7.5)

D) (10.2, 12.6)

I have added the picture showing the traced map onto a coordinate plane to find exact location of treasure.

The coordinates of Tree are (16,21)

Coordinates of rock are (3,2)

Step-by-step explanation:

Let,

Coordinates of treasure be (a,b)

$d_{1}$ = distance from tree to treasure

$d_{2}$ =distance from rock to treasure

$d_{1}=\sqrt{(16-x)^2 + (21-y)^2}$

$d_{2}=\sqrt{x\2 + (y-2)^2}$

Given ratio between rock and tree, $\frac{d_{2}}{ d_{1}}=\frac{5}{9}$ = 0.55_______(Equation.1)

which will be used to locate the treasure.

Now we just need to cross check by putting the coordinates given in the options one by one to find out value of $d_{1}$ , $d_{2}$ and checking if it satisfies the Equation 1.

Check (A) (11.4, 14.2)

$d_{2}$ = 14.8, $d_{1}$ = 8.2,

$\frac{d_{2}}{ d_{1}}$ = 1.8

Check (C) (5.7, 7.5)

$d_{2}$ = 6.13, $d_{1}$ = 16.98,

$\frac{d_{2}}{ d_{1}}$ = 0.36

Check (D) (10.2, 12.6)

$d_{2}$ = 12.8, $d_{1}$ = 10.2,

$\frac{d_{2}}{ d_{1}}$ = 1.25

Check (B) (7.6, 8.8)

$d_{2}$ = 8.2, $d_{1}$ = 14.8,

$\frac{d_{2}}{ d_{1}}$ = 0.55

Which satisfies Equation 1, such that ratio between rock and tree is 5:9 or $\frac{d_{2}}{ d_{1}}=\frac{5}{9}$ = 0.55

So, the coordinates of the treasure are (B) (7.6, 8.8)

7 0

2 years ago

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