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2 years ago

Convert to rectangular coordinates. use exact values. 7 sqrt 2, 135 degrees

Mathematics

1 answer:

Free_Kalibri [48]2 years ago

7 0

Tan 135 = -1

so rectangular coordinates are (-7 sqrt2, 7 sqrt2)

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One brand of cookies is sold in three different box sizes: small, medium, and large. The small box has half the volume of the me

lozanna [386]

Answer:

Step-by-step explanation:

6 0

2 years ago

Mr.Foote placed a cement walk along the side of his house. The walk if 18 inches wide and 32 feet long. What was the cost of the

Sonja [21]

we know that

1 ft-------> is equal to 12 in

step 1

find the area of the walk

18 in wide--------> convert to ft

18/12=1.5 ft

area=1.5*32------> 48 ft²

step 2

find the cost at $2.10 a square foot

multiply 48 ft² by $2.10

48*2.1=$100.80

therefore

the answer is

$100.80

5 0

2 years ago

Ralph is paid twice per month. His work offers a flexible-spending account to be used for medical expenses. This plan allows Ral

slava [35]

In this question, we're assuming Ralph had 0 in his saving funds to start with.

The glasses cost 1200 dollars.

There are 12 months in a year.

Each month, Ralph is given his check two times, that means that Ralph receives his check 2(12), or 24, times per year.

Divide 1200 (the cost of the glasses) by 24 (the amount of times Ralph is given his check)

1200 / 24
Divide both numerator and denominator by 12
100 / 2
Simplify
50

Ralph should have put 50 dollars per pay check to pay for the glasses.

A is your answer.

Have an awesome day! :)

6 0

2 years ago

find the probability that a randomly selected automobile tire has a tread life between 42000 and 46000 miles

maria [59]

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:
$P(X \ \textgreater \ a) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
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 of more than 50,000 miles is given by:
$P(X \ \textgreater \ 50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\ \\ =1-0.98294=\bold{0.0171}$

Part C:

Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
$P(X \ \textless \ a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$
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 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:
$P(X \ \textgreater \ x) =0.06 \\ \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\ \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\ \\ \Rightarrow x=3,669.8+45,000=48,669.8$
Therefore, the value of x is 48,669.8

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

Mustafa, Heloise, and Gia have written more than a combined total of 222222 articles for the school newspaper. Heloise has writt

baherus [9]

Answer:

The Inequality For determining number of equation written by Mustafa for school paper is $x+\frac{1}{4}x+ \frac{3}{2}x\geq 22$ .

Mustafa has written more than 8 articles.

Step-by-step explanation:

Given:

Combined Total Number of articles = 22

Let the number of articles written by Mustafa be 'x'.

Now Given:

Heloise has written $\frac{1}{4}$ as many articles as Mustafa has.

Number of article written by Heloise = $\frac{1}{4}x$

Gia has written $\frac{3}{2}$ as many articles as Mustafa has.

Number of article written by Gia = $\frac{3}{2}x$

Now we know that;

The sum of number of articles written by Mustafa and Number of article written by Heloise and Number of article written by Gia is greater than or equal to Combined Total Number of articles.

framing in equation form we get;

$x+\frac{1}{4}x+ \frac{3}{2}x\geq 22$

Hence the Inequality For determining number of equation written by Mustafa for school paper is $x+\frac{1}{4}x+ \frac{3}{2}x\geq 22$ .

Now Solving the Inequality we get;

Taking LCM for making the denominator common we get:

$\frac{x\times 4}{4}+\frac{1\times1}{4\times1}x+ \frac{3\times2}{2\times2}x\geq 22\\\\\frac{4x}{4}+ \frac{x}{4}+\frac{6x}{4}\geq 22\\\\\frac{4x+x+6x}{4} \geq 22\\\\11x\geq 22\times4\\\\11x\geq 88\\\\x\geq \frac{88}{11} \\\\x\geq 8$

Hence Mustafa has written more than 8 articles.

5 0

2 years ago