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WARRIOR [948]
1 year ago
14

There is a 20% chance that you will be elected president of the history club. There is a 20% chance that your friend will be ele

cted treasurer of
the history club. You use a random number generator to randomly generate 50 numbers from 0 to 99. The results are shown in the table
below. The digits 1 and 2 in the tens place represent you being elected president. The digits 1 and 2 in the ones place represents your friend
being elected treasurer. What is the experimental probability that you are elected president and your friend is elected treasurer? Write your
answer as a fraction in simplest form
10 73 97 53 20 36 57 76 61 73
46 1371 43 94 4 11 6220 33
70 6 15 39 22 94 7 32 36 22
11 25 31 57 65 50 31
13 45 29 83 60 12 26 33 33
The experimental probability is
Mathematics
1 answer:
RideAnS [48]1 year ago
8 0

Answer:

Both of you have a 50/50 (1) chance to be elected

Step-by-step explanation:

Depending onthe independant variables of the understanding of the hipotnus it should be 50/50 or 1

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Gavin and Jack are practicing shots against their goalie. On their last 15 attempts, Gavin made 6 and Jack made 7. Based on this
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Answer:

\dfrac{14}{75}

Step-by-step explanation:

Gavin made 6 out of 15 shots, so the probability that Gavin's next shot will be  successful is

\dfrac{6}{15}=\dfrac{2}{5}

Jack made 7 out of 15 shots, so the probability that Jack's next shot will be  successful is

\dfrac{7}{15}

The probability that they both make their next shot successfully is

\dfrac{2}{5}\cdot \dfrac{7}{15}=\dfrac{14}{75}

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2 years ago
The formula h -15 = 3.2t gives the height h in inches of a plant t weeks after planting.
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Substitute t = 0, to get the initial height
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2 years ago
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2. Santiago wants the lateral surfaces of the
MakcuM [25]

Answer:

262 square feet  of the ramp will be painted red.

Step-by-step explanation:

The lateral surface is surface area of the sides of the ramp without including the top and bottom faces

The ramp shown has three surfaces

The 2 sides are triangles with length 20 and height 8.5

The back is a rectangle shape with length 12 and height 8.5

Step 1: Finding the  Area of triangle

The area of the triangle  = \frac{1}{2} length \times height

The area of the triangle  = \frac{1}{2} (20 \times 8.5)

The area of the triangle  = \frac{170}{2}    

The area of the triangle = 85 square feet

Area of 2 triangles = 85 \times 2  =  170

Step 2: Finding the  Area of Rectangle

Area of rectangle = length  \times height

so,

Area of back rectangle = 12  \times 8.5 = 102  square feet

Step 3: Finding the total lateral surface area

Total Lateral Surface Area

= Area of triangle + Area of Rectangle

= 272 square feet

8 0
1 year ago
The slope-intercept form of a linear equation is y = mx + b, where x and y are coordinates of an ordered pair, m is the slope of
Serjik [45]
Y = mx + b
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Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}

\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

\frac{\sin(x+y)}{\cos(x+y)} by quotient identity for tangent

\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}

-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}

-\frac{8-2\sqrt{12}}{4}

There is a perfect square in 12, 4.

-\frac{8-2\sqrt{4}\sqrt{3}}{4}

-\frac{8-2(2)\sqrt{3}}{4}

-\frac{8-4\sqrt{3}}{4}

Divide top and bottom by 4 to reduce fraction:

-\frac{2-\sqrt{3}}{1}

-(2-\sqrt{3})

Distribute:

\sqrt{3}-2

6 0
1 year ago
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