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2 years ago

8

Using a sixteen-sided number cube, what is the probability that you will roll an even number or an odd prime number? The number

1 isn't an odd prime. Round to three decimals.

1 answer:

grandymaker [24]2 years ago

8 0

1. Let a: rolling an even number, b: rolling an odd prime number

2. a={2, 4, 6, 8, 10, 12, 14, 16}, b={3, 5, 7, 11, 13}

3. So n(a)=8, n(b)=5 , n(Sample Space)=n(S)=16

4. Since a and b are mutually exclusive events, that means they have no common elements, P(a or b)=P(a)+P(b)= $\frac{n(a)}{n(S)}+ \frac{n(b)}{n(S)}= \frac{8}{16}+ \frac{5}{16}= \frac{13}{16}= 0.813$

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1. Your furniture store sells two types of dining room tables. The first, type A, costs $265 and you make a $25 profit on each o

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Let x represent the number of type A table and y represent the number of type B tables.
Minimize: C = 265x + 100y
Subject to: x + y ≤ 40
25x + 13y ≥ 760
x ≥ 1, y ≥ 1

From, the graph the corner points are (20, 20), (39, 1), (30, 1)
For (20, 20): C = 265(20) + 100(20) = $7,300
For (39, 1): C = 265(39) + 100 = $10,435
For (30, 1): C = 265(30) + 100 = $8,050

Therefore, for minimum cost, 20 of type A and 20 of type B should be ordered.

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2 years ago

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Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an

Charra [1.4K]

1. Consider square ABCD. You know that

$A_{ABCD}=AD^2=200,$

then

$AB=BC=CD=AD=10\sqrt{2}.$

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

$FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.$

3. Consider trapezoid BFGC. Its area is

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot h,$ where h is the height of trapezoid and is equal to half of AB. Thus,

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.$

4.

$A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.$

5. Note that angles EGC and CGD are supplementary and

$\sin \angle CGD=\sin \angle EGC.$

Then

$A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.$

Answer: $A_{CGD}=41.$

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2 years ago

Mr.Anderson is building a triangular- shaped roof for his shed. The triangle must be isosceles with its base (noncongruent) side

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The side of the congruent side of the triangle should be greater than 7 feet.

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2 years ago

A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed, t

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If we let
x as the distance traveled by the boat
y as the distance between the boat and the lighthouse.

Then, we have:
tan 18°33' = 200 / (x + y)
and
tan 51°33' = 200 / y

Solving for y in the second equation:
y = 200 / tan 51°33'

Rearranging the first equation and substituting y
x = 200 / tan 18°33' - 200 / tan 55°33'
x = 458.81 ft

Therefore, the boat traveled 458.81 ft before it stopped.

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2 years ago

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Imogen sees a design for a garden in a magazine. She changes the scale because she wants a smaller garden. The figure below show

faust18 [17]

Answer:

Length A of the magazine garden is 28 feet is true.

Length B of the Magazine garden is 17.5 feet is true.

Length C of the Magazine garden is 35 feet is true.

Step-by-step explanation:

Please refer the attached figure.

Given:-

Magazine Scale is 1 inch = 3.5 feet.

Imogen's scale is 2 inches = 6 feet,

Therefore Imogen's scale is 1 inch = 3 feet.

In Triangle ABC,

side a = 8 inches, side b= 5 inches, side c= 10 inches.

To find:- The correct 3 options(actual size of sides of triangle ABC).

Solution:-

In triangle ABC,

Let,

AB=side a =8 inches

BC=side b =5 inches

AC=side c= 10 inches

Now, as per the given scale's of Magazine and Imogen's we can find the actual sizes.

Length A as per Magazine scale (1 inch = 3.5 feet) is,

$\therefore Length\ A=8\times 3.5$ -------------(as per scale)

$\therefore Length A=28\ feet$

and Length A as per Imogen's scale(1 inch =3 feet) is,

AB = side a

$\therefore Length\ A=8\times 3$ -------------(as per scale)

$\therefore Length A=24\ feet$

Therefore Length A of the magazine garden is 28 feet is true.

Now, Length B as per Magazine scale (1 inch = 3.5 feet) is,

BC = side b

$\therefore Length\ B=5\times 3.5$

$\therefore Length B=17.5\ feet$

Length B as per Imogen's scale (1 inch = 3 feet) is,

BC = side b

$\therefore Length\ B=5\times 3$

$\therefore Length B=15\ feet$

Therefore Length B of the Magazine garden is 17.5 feet is true.

Now, Length C as per Magazine scale (1 inch = 3.5 feet) is,

AC = side c

$\therefore Length\ C=10\times 3.5$

$\therefore Length C=35\ feet$

Length B as per Imogen's scale (1 inch = 3 feet) is,

AC = side c

$\therefore Length\ C=10\times 3$

$\therefore Length B=30\ feet$

Therefore Length C of the Magazine garden is 35 feet is true.

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2 years ago

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