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3 years ago

How do I set up this problem?

Mathematics

1 answer:

Margaret [11]3 years ago

4 0

Hey there!

We know that perimeter is all the sides added up.

70= 2w+2L <== this is the basic equation

We know that the rectangles length is three times as long as it is wide.

So, you know that is is three times the length of the width. So, put 3 times w in as length and times that by two since there are two length sides, and keep width the same. 2w because there are two width sides. Then, solve.

70=2(3w)+2w

70= 6w+2w

70=8w

w=8.75

We can check this by plugging it in to our first equation and seeing if it works. First, we need to multiply 8.75 by 3 to get the value for length.

8.75*3= 26.25

2(26.25)+2(8.75)= 70 <=== it works!

So, the length is 26.25 cm and the width is 8.75 cm.

I hope this helps!

~kaikers

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3 years ago

Accuracy in taking orders at a drive-through window is important for fast-food chains. Periodically, QSR Magazine publishes "The

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Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that $p = 0.905$

You and 2 friends:

So 3 people in total, which means that $n = 3$

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412$

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009$

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245$

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

With what we found in b:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991$

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

$p = 0.905*0.095*0.095 = 0.0082$

0.0082 = 0.82% probability that only your order will be filled correctly

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3 years ago

A random sample of 121 automobiles traveling on an interstate showed an average speed of 73 mph. from past information, it is kn

tatiyna

121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range

$(73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)$

The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.

The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:

$\sigma = \dfrac{ 11}{\sqrt{121}} = 1$

So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range

$(73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75)$

That's probably the answer they're looking for.

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3 years ago

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Let Daniel's age be 'x' years , hence, Lisa's age is (x - 5) years. Let Grandma's age be 'y' = 54 years.
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x(x - 5) = 2/3 × y
x² - 5x = 2 × 54 years / 3
x² - 5x = 36 years
x² - 5x - 36 = 0
∆ = (-5)² - 4(1)(-36) = 25 + 144 = 169
√∆ = √169 = 13
$x\: = \: \frac{ - b \: \frac{ + }{ } \: \sqrt{ {b}^{2} \: - \: 4ac} }{2a}$
x1 = (5+13)/2, x2 = (5-13)/2
x1 = 9, x2 = -4
Since, age can't be negative,
Daniel's age = x = 9 years.

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4 years ago

Plsss helpppp 6x^2+4x-42

melisa1 [442]

Answer:

Step-by-step explanation:

6x^2+4x-42

2(3x^2+2x-21)=0

3x^2+2x-21=0

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3x^2+9x-7x-21=0

3x(X+3)-7(X+3)=0

(3X-7)(X+3)=0

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3 years ago