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miskamm [114]
4 years ago
5

How many outcomes are there in the sample space for tossing 2 coins?

Mathematics
2 answers:
lukranit [14]4 years ago
4 0
Ultimately, one can define one's events and sample space however one wants. But we usually design it with some sort of problem in mind. You can understand it by assuming that 1 coin is red and the other is black. In this case there are exactly 4 outcomes<span> when you flip these 2 coins, that is HH, HT, TH or TT.</span>
Veseljchak [2.6K]4 years ago
4 0

Answer:

There are 4 outcomes

Step-by-step explanation:

cuz 2*2=4

You might be interested in
Using the formula z=Kx/y<br> If z = 6 when x = 3 and y = 4, then what is z when x = 5 and y = 2?
Veseljchak [2.6K]

Answer:

z=kx/y

zy=kx

zy/x=kx/x ( divided both side by x. to find k)

k=zy/x

k=6×4/3(z=6,y=4,x=3)

k=24/3

k=8

so z=kx/y

z=8×5/2(replace the numbers in the place of variables,k=8,x=5,y=2)

z=40/2

z=20

3 0
3 years ago
To encourage bulk buying a company reduces the list price of $9000 per unit by seven cents times the number of units bought writ
Misha Larkins [42]

Answer:

R(x) = 8999.93x

Step-by-step explanation:

The original price is $9000 per unit. The unit is x, so if you buy x units, you pay 9000x.

The original price function is

R(x) = 9000x

The discount is 7 cents per unit bought, so if you buy x units, the discount is 9x in cents, or 0.09x in dollars. This discount is subtracted from the original price, so the discounted price is

R(x) = 9000x - 0.07x

R(x) = 8999.93x

Answer: R(x) = 8999.93x

7 0
3 years ago
1. (04.06A) What is the slope of the line shown in the graph? (4 points) −1 −2 - 1/2 2
never [62]

Answer:

The best and most correct answer among the choices provided by your question is the the first choice. We can conclude from the figure you specified that the "Points are graphed at negative 3 comma 2 and negative 1 comma negative 1". I hope my answer has come to your help.

Step-by-step explanation:

A: −3 over 2

6 0
3 years ago
Simplify each expression to a single trig function or number cosx(secx-cosx)
AlladinOne [14]
I did this test  b4, yours is answer #number 12

Convert things to their basic forms. 
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>

<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>

<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>

<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>

<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>

<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>

<span>4. 1- cos^2θ </span>
<span>sin^2 </span>

<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>

<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>

<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>

<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>


<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>

<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>

<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>


<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>

<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>

<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
5 0
3 years ago
HELP ASAP Solve log7 b &gt; 2. Question 18 options: b &gt; 7 b &gt; 49 b &gt; 2–7 b &gt; 14
labwork [276]

Answer: Second Option

b > 49

Step-by-step explanation:

We have the following expression:

log_7(b) > 2

We have the following expression:

To solve the expression, apply the inverse of log_7 on both sides of the equality.

Remember that:b ^ {log_b (x)} = x

So we have to:

7^{log_7(b)} > 7^2

b > 7^2

b > 49

The answer is the second option

3 0
4 years ago
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