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Tanya [424]
4 years ago
5

What is :

Mathematics
1 answer:
Natasha2012 [34]4 years ago
6 0
Well if they match 60% of the $8,000 then it will be $14,300 but if they match 60% of the 8,000+1,500 then it would be $15,200.
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The mean of the following numbers is 15. 14,10,18,,21,15,14​
Leona [35]

Answer:

= 15.285714285714

Step-by-step explanation:

6 0
3 years ago
Find the standard deviation of the distribution in the following situations. (Round your answers to two decimal places.) (a) MEN
Arturiano [62]

Answer:

a) 15.579

b) 39.019

Step-by-step explanation:

a) For the top 2% the z-value from the z table

z = 2.054

thus,

z = \frac{\textup{X-Mean}}{\sigma}

here

X = 132

Mean = 100

thus,

2.054 = \frac{\textup{132-100}}{\sigma}

or

σ = 15.579

b) Using tables, we get for Z = 0.89 the value of 0.8133 and for Z= 0.90 the value of 0.8159

By interpolation, values for 0.815 and we get Z=0.897

thus,

z = \frac{\textup{X-Mean}}{\sigma}

here

X = 220

Mean = 185

thus,

0.897 = \frac{\textup{220-185}}{\sigma}

or

σ = 39.019

6 0
3 years ago
NEED HELP ASAP PLEASE
IrinaVladis [17]
Z would also be 63 degrees. Triangle WXZ is isosceles, as sides WX and WZ are marked as congruent. This means that the two base angles of the triangle (the bottom two) are congruent. You already know that one of the base angles, X (or WXY), is 63 degrees, so using this reasoning, the other base angle, YZW, is 63 degrees as well. I hope this helps!
8 0
4 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
Solve for all solutions (2x+3)2=81
Scrat [10]
X=12
Hope this helps

3 0
3 years ago
Read 2 more answers
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