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9966 [12]
3 years ago
14

The coordinate plane below represents a city. Points A through F are schools in the city.

Mathematics
1 answer:
hodyreva [135]3 years ago
6 0
PART A

Refer to the diagram below for the region that only has point C and F inside the shaded region

The three inequalities can be:
y ≥ 1
x ≥ 1
y ≤ -x + 8
---------------------------------------------------------------------------------------------------------

PART B

Coordinate of C is (2, 2)
Coordinate of F is (3, 4)

For the inequality y ≥ 1, the y-coordinate of both point C and F are more than 1

For the inequality x ≥ 1, the x-coordinate of both point C and F are more than 1

For the inequality y ≤ -x + 8
Start with point C, we substitute the x-coordinate = 2 and check whether -x + 8 is indeed more than y
-x + 8 = -(2) + 8 = 6 and 6 ≥ 2

Point D (3, 4), substituting x = 3
-x + 8 = -(3) + 8 = 5 and 5 ≥ 4

So point C and F are solutions to the inequalities
--------------------------------------------------------------------------------------------------------------

PART C

Refer to the second diagram below, the points below the line -2x + 2 are A., B, and D. We choose the point below the line because the inequality want 'y' to be less than -2x + 2


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Step-6.95 gallons in 1 month


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Step-by-step explanation:

  • You can use the distributive formula to answer:

a(b - c) = ab - ac

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2 years ago
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Nathan bought a cd for his friend. The length and width of the cd case is 4 18⁄25 inches. What is the area of the cd case?
ale4655 [162]

Answer:

Hence the area of CD bought by Nathan is <em>22.2784 square inches.</em>

Step-by-step explanation:

The length and width of a CD bought by Nathan is 4\frac{18}{25}=\dfrac{118}{25} inches.

As CD have the same length and width this means it will be of the shape of a square.

Hence, the area of CD is given by:

Area of CD=length×width

                 = 4\dfrac{18}{25}\times4\dfrac{18}{25}

                 =\dfrac{118}{25}\times\dfrac{118}{25}

                 =\dfrac{118\times118}{25\times25}

                 =\dfrac{13924}{625}

                 =22.2784

Hence the area of CD bought by Nathan is 22.2784 square inches.


3 0
3 years ago
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0
3 years ago
Find the amount accumulated after investing a principle P for T years at an interest rate compounded k times per year.
BartSMP [9]
1. 3500(1+ \frac{.05}{4}) ^{4*10}
3500(1+.0125) ^{40}
3500(1.64) = 5740

2. 25300(1+ \frac{.045}{12} ) ^{12*25}
25300(1+.00375) ^{300}
25300(3.07) = 77671

Hope that helps.  Feel free to ask any questions
8 0
3 years ago
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