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Molodets [167]
3 years ago
11

Onna received a $80 gift card for a coffee store. She used it in buying some coffee that cost $8.09 per pound. After buying the

coffee, she had 39.55 left on her card. How many pounds of coffee did she buy?
Mathematics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

She bought 5 pounds

Step-by-step explanation:

Original worth of gift card = $80

Balance on card after purchase = $39.55

Amount used for purchase = 80 - 39.55 = $40.45

Next let us determine how many pounds of coffee she bought as follows:

Price of coffee = $8.09 per pound

amount spent on coffee = $40.45

Lett the number of pounds of coffee she bought be x

$ 8.09 = 1 pound

$40.45 = x pounds

since $8.09 = 1 pound

\$\ 1 = \frac{1}{8.09}\ pounds

\$\ 40.45 = 40.45 \times\ \frac{1}{8.09} \\= \frac{40.45}{8.09}\\ = 5\ pounds

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Thinking about the area of a standard normal distribution , as z values increase, do the areas to the left of z increase, decrea
Serggg [28]
Refer to a diagram of the normal distribution shown below.
As the z-value increases to the right, areas to the left of z (shown shaded) will increase.

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As z-values increase, areas to the left of z increase.

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4 years ago
Helpppp pleaseeeeeeeeeeeeeee
Vesna [10]

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Find below the calculations of the two areas, each with two methods. The results are:

  • Upper triangle:

                     Area=5000\sqrt{3}units^2

  • Lower triangle:

                     Area=14,530m^2

Explanation:

<u>A) Method 1</u>

When you are not given the height, but you are given two sides and the included angle between the two sides, you can use this formula:

           Area=side_1\times side_2\times sin(\alpha)

Where, \alpha is the measure of the included angle.

1. <u>Upper triangle:</u>

          side_1=200units\\ \\ side_2=100units\\ \\ \alpha =60\º\\ \\ Area=200units\times 100units\times sin(60\º)/2\\ \\ Area=5000\sqrt{3}units^2

2. <u>Lower triangle:</u>

         side_1=231m\\ \\ side_2=150m\\ \\ \alpha =123\º\\ \\ Area=231m\times 150m\times sin(123\º)/2\\ \\ Area=14,529.96m^2\approx14,530m^2

<u></u>

<u>B) Method 2</u>

You can find the height of the triangle using trigonometric properties, and then use the very well known formula:

            Area=(1/2)\times base\times height

Use it for both triangles.

3. <u>Upper triangle:</u>

The trigonometric ratio that you can use is:

                    sine(\alpha)=opposite\text{ }leg/hypotenuse

Notice the height is the opposite leg to the angle of 60º, and the side that measures 100 units is the hypotenuse of that right triangle. Then:

         sin(60\º)=height/100units\\ \\ height=sin(60\º)\times100units\\ \\ height=50\sqrt{3}units

Area=(1/2)\times base\times height=(1/2)\times 200units\times 50\sqrt{3}units=5,000\sqrt{3}units^2

3. <u>Lower triangle:</u>

<u />

         sin(180\º-123\º)=height/231m\\ \\ height=sin(57\º)\times 231m\\ \\ height=193.7329m^2<u />

<u />

<u />Area=(1/2)\times base\times height=(1/2)\times 150m\times 193.7329m^2\\\\  Area=14,529.96m^2\approx 14,530m^2<u />

5 0
3 years ago
Point A is at (2,-8) and the point C is at (-4,7). Find the coordinates of point B on AC such that the ratio of AB to BC is 2:1.
pentagon [3]

Answer:

The coordinates of point B are (-2, 2)

Step-by-step explanation:

We have two points: A and C.

The coordinates for A are (2, -8) and the coordinates for C are (-4, 7).

We have to find the coordinates of the point B, that satisfies the condition that the distance AB is 2 times the distance BC.

We also know that B is a point of the line AC.

We can calculate the line AC as a linear function y=mx+b.

The slope m is:

m=\dfrac{y_c-y_a}{x_c-x_a}=\dfrac{7-(-8)}{-4-2}=\dfrac{15}{-6}=-2.5

Then, the y-intercept b can be calculated using the coordinates of one of the points, in this case point A:

y=-2.5x+b\\\\b=y_a+2.5x_a=-8+2.5*2=-8+5=-3

Then, we know that B is a point of the linear function y=-2.5x-3, within the range x ∈ (-4; 2).

To have a ratio AB to BC of 2 to 1, we can divide the length of the line AC in 3 parts, and the point B will be located in the  end of the segment nearer to point C.

In the picture attached, you can see the division of the segment AC in three parts and the location of point B=(x, y).

Applying the Thales theorem, we can divide the segment in the y-axis in three and calculate y, and the same for the x-axis.

Then, the coordinate y for the point B is:

y=y_c-(y_c-y_a)/3\\\\y=7-[7-(-8)]/3=7-15/3=7-5=2\\\\\\x=x_c-(x_c-x_a)/3\\\\x=-4-(-4-2)/3=-4-(-6)/3=-4+2=-2

Then, the point B has coordinates (-2, 2).

We can verify the distances as:

AB=\sqrt{(2-(-2))^2+((-8)-2)^2}=\sqrt{16+100}=\sqrt{116}\\\\\\BC=\sqrt{((-2)-(-4))^2+(2-7)^2}=\sqrt{4+25}=\sqrt{29}\\\\\\\dfrac{AB}{BC}=\dfrac{\sqrt{116}}{\sqrt{29}}=\sqrt{\dfrac{116}{29}}=\sqrt{4}=2

4 0
3 years ago
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