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3 years ago

Help thanks!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

irinina [24]3 years ago

6 0

The answer is C)(0.38, -2.85) and (2.62, 3.85)
Hope this helps

alina1380 [7]3 years ago

4 0

X²-3=3x-4
x²-3x+1=0
x=(3+/-√(9-4))/2=(3+/-2.24)/2
x1=2.62
x2=0.38
C)

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Im going to give 15 points for this question please answer it

givi [52]

Area of the rectangle plus the area of the parallelogram...

53.3(11.1+6.8)

53.3(17.9)

954.07 in^2

6 0

3 years ago

Need help please... In gym class, you run 3/5 mile. Your coach runs 10 times that distance each day. How far does your coach run

vovangra [49]

3/5 x 10= 6 miles because coach runs 10 times what you run

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3 years ago

Which equation are equivalent to 1/5+2/3|2-x|=4/15?

disa [49]

Answer:

• 2/3|2 -x| = 1/15

• |2 -x| = 1/10

• 1/5 +|4/3 -2/3x| = 4/15

Step-by-step explanation:

Starting with the given equation, subtract 1/5 = 3/15.

(2/3)|2 -x| = 4/15 - 3/15

(2/3)|2 -x| = 1/15 . . . . . . . . matches the first choice

Now, multiply by 3/2.

|2 -x| = 3/30

|2 -x| = 1/10 . . . . . . . . . . . . matches the third choice

___

If you decide to distribute the coefficient 2/3 instead, you have

1/5 + |2·2/3 -x·2/3| = 4/15

1/5 + |4/3 -2/3x| = 4/15 . . . . matches the selected choice

5 0

3 years ago

Read 2 more answers

A woman in a highland village in the Andes knits sweaters and sells them for export. She also takes care of her family and helps

Mnenie [13.5K]

Answer:

Expected number of sweaters per month can be given as follows:

E(X) = Σ x P(X = x)

Now,

E(X) = [2 * 0.1 + 3* 0.1+ 4* 0.2 + 5* 0.3 + 6* 0.2 + 7 * 0.1]

E(X) = 4.7.

Var(X) = E(X^2) – [E(X)]^2

We have E(X) = 4.7. Thus, [E(X)]2= 4.7*4.7 = 22.09.

Now E(X^2) = [2*2 * 0.1 + 3*3* 0.1+ 4*4* 0.2 + 5*5* 0.3 + 6*6* 0.2 + 7*7*0.1]

E(X^2) = 24.1

Thus by formula, Var(X) = E(X2) – [E(X)]2

Var(X) = 24.1-22.09

Var(X) = 2.01

Given that exporter pays the $12 for each sweater. The woman pays $2 per sweater. The cost of shipment is $3 irrespective of the number of sweaters. Now, let m is the number of sweaters she made. Thus, the total cost she would have to pay would be

Total cost by woman = 2m+3

The total cost paid by the exporter would be = 12m.

Now the profit of woman would be given by,

= The total cost exporter pay – cost paid by the woman

= 12m – (2m +3)

= 12m – 2m -3

= 10m – 3.

Now expected profit made by the woman is given in the following table below:

E(Profit) = Σ profit* P(X = x)

In a similar way, as we have done in part (a).

E(Profit) = [17 * 0.1 + 27* 0.1+ 37* 0.2 + 47* 0.3 + 57* 0.2 + 67 * 0.1]

E(Profit) = 44.

Now, we calculate the variance:

Var(profit) = E(profit^2) – [E(profit)]^2

Var(profit) =

E(profit^2) = [17*17 * 0.1 + 27*27* 0.1+ 37*37* 0.2 + 47*47* 0.3 + 57*57* 0.2 + 67*67*0.1]

E(profit^2) = 2137.

[E(profit)]^2 = 44*44 = 1936.

Thus, the variance can be given as =

Var(profit)= 2137 – 1936

Var(profit) = 201.

6 0

3 years ago

In a particular course, it was determined that only 70% of the students attend class on Fridays. From past data it was noted tha

siniylev [52]

Answer: Probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

Step-by-step explanation:

Since we have given that

Probability that students attend class on Fridays = 70% = 0.7

Probability that who went to class on Fridays would pass the course = 95% = 0.95

Probability that who did not go to class on Fridays would passed the course = 10% = 0.10

Let A be the event students passed the course.

Let E be the event that students attend the class on Fridays.

Let F be the event that students who did not attend the class on Fridays.

Here, P(E) = 0.70 and P(F) = 1-0.70 = 0.30

P(A|E) = 0.95, P(A|F) = 0.10

We need to find the probability that they did not attend on Fridays.

We would use "Bayes theorem":

$P(F\mid A)=\dfrac{P(F).P(A\mid F)}{P(E).P(A\mid E)+P(F).P(A\mid F)}\\\\P(F\mid A)=\dfrac{0.30\times 0.10}{0.70\times 0.95+0.30\times 0.10}\\\\P(F\mid A)=\dfrac{0.03}{0.695}=0.043$

Hence, probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

8 0

3 years ago