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Natasha_Volkova [10]
3 years ago
11

COuld someone help em with this :)

Mathematics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:

 by m x equals #33

Step-by-step explanation:

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A television normally costs 1420.Due to a store sale the television is now 1,022.40.What was the percentage taken off the televi
Zielflug [23.3K]

Cost of Television = $1420

Cost of Television after store sale =$1022.40

Drop in price= 1420-1022.40= 397.60.

Price taken off = $ 397.60.

Percentage taken off= \frac{Difference }{original} x 100.

                                 = \frac{397.60}{1420} x 100=28%.

Percent off = 28%.

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3 years ago
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5. Using the information from problem 4, if the club sells 40 tickets in advance, what is the least number the club needs to sel
larisa [96]

Answer:

8

Step-by-step explanation:

40 + 5 (at the door)=45

(president's goal) 400 divided by 45= approximately 8

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3 years ago
Ashley got a haircut at the local hair salon. The total for her services was $95 and she will give the stylist a 30% tip, what i
Temka [501]

Answer:

divide 95÷30% then that will be your answer

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3 years ago
What number would X be in 5 + (x - 2) = -8
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-11

Step-by-step explanation:

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3 years ago
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(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a
Fynjy0 [20]

Step-by-step explanation:

Given that,

a)

X ~ Bernoulli (p_x) and Y ~ Bernoulli (y_x)

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

b)

If X and Y can not be both equal to  1

then,

p_z = P(X=1 or Y=1)\\

p_z = P(X=1)+P(Y=1)-P(=1 and Y =1)

p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y

c)

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a  Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0
3 years ago
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