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coldgirl [10]
3 years ago
6

What is 1 and 3/4 as a mixed number

Mathematics
2 answers:
vazorg [7]3 years ago
4 0
The answer is 7/4               ,
Wittaler [7]3 years ago
3 0
7/4 I believe

Gr8 Day M8
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Nick's bank account balance was $21. if he writes a check for $30 and his bank charges him $35 for overdrawing his account, find
natulia [17]
It is saying that the Balance was 21 if he wrote a check for 30 and The bank was charging him $35 for overdrawing his account of basically you’re gonna just Add so 21+30+35 is 86
4 0
3 years ago
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I need help asp . I'm running out of time​
frutty [35]

Answer:

A = 104 ft²

Step-by-step explanation:

We're given that the width of the rectangle is 8 so that's the value of w. All we have to do is just plug in 8 into the equation:

A=w²+5w

A=(8)²+5(8)

A=64+40

A = 104 ft²

7 0
3 years ago
What does "tmi" mean ?
Stella [2.4K]

Answer:

To much info

Step-by-step explanation:

6 0
3 years ago
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$12,000 is invested for 4 years at a simple interest rate of 1.5%. How much does the investment earn?
bixtya [17]
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5 0
3 years ago
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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