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Strike441 [17]
3 years ago
13

James earns $584 per week. If he receives a pay increase of 7.5%, find:

Mathematics
2 answers:
dsp733 years ago
5 0
584 times 7.5% = 584 times 0.075= 43.8
A: 43.8 dollars

liberstina [14]3 years ago
3 0
So the pay after the increase is 4380 so subtract 4380-584 which equals 3796.
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HELPPPP<br> What is the exact area for this figure?
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8 cubic meters squared.

Step-by-step explanation:

2*4=8

8/2=4

Since they're the same... You would multiply the product by two.

4*2=8

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3 years ago
The cost of highlighters is proportional to the number of highlighters purchased. If a 4-pack of highlighters costs $2.40, what
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$0.60 per highlighter

Step-by-step explanation:

You just do $2.40/4 and you get $0.60

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Graph each inequality and graph its solution.<br> 3) n - 11 &gt; -21
Vera_Pavlovna [14]
The correct answer is
Inequality: n>-10
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7 0
3 years ago
Please help !!! What’s the answer to this??
Aleksandr [31]

Answer:

Average speed is distance/time. You could say, 12/6 = 2 miles per second and say that’s the answer, but we relate to mph so much better. Since there are 60 x 60 = 3600 sec in an hour, that same speed is:

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No “car” can travel this fast. Some rocket cars have gone about 1/10th this speed.

7 0
2 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
Read 2 more answers
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