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2 years ago

A belt and a wallet cost $42 while 7 belts and 4 wallets cost $213. Calculate the cost of each item.

Mathematics

2 answers:

irina1246 [14]2 years ago

6 0

4 pair × $42 = 168
3 belts cost 45 and one will cost 15
if belt costs 15 wallet will cost $27

Fittoniya [83]2 years ago

3 0

Let, the cost of a belt = x
cost of a wallet = y

Then, system of equations would be:
x + y = 42
7x + 4y = 213

Multiply 1st equation by 4,
4x + 4y = 168
Substitute it from 2nd equation,
3x = 45
x = 15

Now, substitute it in 1st equation,
15 + y = 42
y = 42 - 15 = 27

In short, Belt costs $15 and wallet costs $27

Hope this helps!

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Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.

Part A:

Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
$P(a \ \textless \ X \ \textless \ b) = P(X \ \textless \ b) - P(X \ \textless \ a) \\ \\ P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
$P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\ \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}$

b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
$P(X \ \textgreater \ a) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
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The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
$P(X \ \textgreater \ 50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\ \\ =1-0.98294=\bold{0.0171}$

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Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
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$P(X \ \textless \ a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
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The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
$P(X \ \textless \ 38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\ \\ =P(z\ \textless \ -2.966)=\bold{0.0015}$

d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
$P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.

Thus:
$P(X \ \textgreater \ x) =0.06 \\ \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\ \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\ \\ \Rightarrow x=3,669.8+45,000=48,669.8$
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e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
$P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.

Thus:
$P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56$
Therefore, the value of x is 40,152.56

4 0

2 years ago