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NeX [460]
3 years ago
11

Solve for the missing side of this right triangle. 24 mi 16 mi.

Mathematics
1 answer:
Over [174]3 years ago
3 0

Answer:

384

Step-by-step explanation:

I just went to my calculator and did 24×16 and I got 384

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Solve the inequality for the missing variable<br> 24 -1t<br> 1/5
Scorpion4ik [409]

Answer:

−t+24

Step-by-step explanation:

8 0
3 years ago
7) Using the Distributive Property (think of the double rainbows!), write an equivalent expression to 11(7 - 3). Your answer sho
maxonik [38]

Answer:

77-33

Step-by-step explanation:

1)

11(7-3)

2)

(11 x 7) - (11 x 3)

3)

77 - 33

Answer:

77-33

6 0
3 years ago
2 2/3 + 4 1/8 reduced to lowest terms
Olegator [25]

Answer:

6 19/24

Step-by-step explanation:

2 2/3 + 4 1/8

3 times 8 to find the common denominator which then gives you

2 16/24 + 4 3/24

That equals

6 19/24

Mark me as brainliest if this helps!

3 0
4 years ago
Read 2 more answers
A player tosses a die 6 times.If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the ex
swat32

Answer:

E(x)=-0.2101

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x_1*p(x_1)+x_2*p(x_2)

Where x_1 and x_2 are the values that the variable can take and p(x_1) and p(x_2) are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that x_1 is equal to 2 and x_2 is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

P(a)=nCa*p^a*(1-p)^{n-a}

Where nCa=\frac{n!}{a!(n-a)!}

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability  p(x_1) that a player get at least two times number 6, is calculated as:

p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633

On the other hand, the probability p(x_2) that a player don't get  at least two times number 6, is calculated as:

p(x_2)=1-p(x_1)=1-0.2633=0.7367

Finally, the expected value of the amount that the player wins is:

E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101

It means that he can expect to loses 0.2101 dollars.

3 0
3 years ago
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The square of the product of 4 and a number.
aleksley [76]

the square root of 4x


\sqrt{4x}

8 0
3 years ago
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