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GalinKa [24]
3 years ago
6

Pls help math problem

Mathematics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

angle plm+angle lpm+angle pml=180°[by angle sum property]

21°+angle lpm+38°=180°

59°+ lpm=180°

lpm=180°-59°

lpm=121°

lpm+npl=180°[by limear pair]

121°+npl=180°

npl=180°-121°

npl=59°

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Six minus a number divided by 3 equals 8. What is the number?
Gre4nikov [31]

Answer:

-18

Step-by-step explanation:

let n be the number

(6-n)/3 = 8

6-n = 24

6-24 = n

-18 = n

5 0
3 years ago
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What is the value of x?<br> 40°<br> (6x + 2)
Anvisha [2.4K]

Answer:

x = 8

Step-by-step explanation:

40 + 6x + 2 = 90   {Complementary angles}

Combine like terms

40+ 2 + 6x = 90

     42 +6x = 90         {Subtract 42 form both sides}

             6x = 90 - 42

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4 years ago
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earnstyle [38]
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3 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
4 years ago
A movie theater determines its profit, in dollars, using the expression 10n - 1000, where n is the number of tickets sold.
Alja [10]

Answer:

E

600

Step-by-step explanation:

10n - 1000 = 5000

10n = 6000

n = 600

HOPE THIS HELPS

PLZZ MARK BRAINLIEST

7 0
3 years ago
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