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Hunter-Best [27]
3 years ago
15

37= -3 + 5 (x+6) solve for x

Mathematics
2 answers:
Tomtit [17]3 years ago
7 0

Answer:

x = -6

Step-by-step explanation:

1. Pull out like factors: 24+4x=4•(x + 6)

2. Solve: 4=0 (This equation has no solution. A a non-zero constant never equals zero.)

3. Solve: x+6=0 (Subtract 6 from both sides of the equation.)

That's how you will get x = -6

Yanka [14]3 years ago
6 0

Answer:

x = 2

Step-by-step explanation:

37= -3 + 5 (x+6)

Distribute 5(x+6)

37= -3 + 5x + 30

Combine like terms (30-3)

37= 5x + 27

Subtract 27 to both sides (37-27)

10 = 5x

Divide 5 to both sides (10/5)

x = 2

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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali
Dominik [7]

Answer:

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

Y(W,W) Y(W,T) Y(W,F) Y(W,S)

Y(T,W) Y(T,T) Y(T,F) Y(T,S)

Y(F,W) Y(F,T) Y(F,F) Y(F,S)

Y(S,W) Y(S,T) Y(S,F) Y(S,S)

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

       = 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

      = (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

      = 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

  =(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

  = 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)

      = (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

       = 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

The PMF plot is attached as a photo here.

7 0
3 years ago
2r+8-r=-7<br> Need help ASAP
Anvisha [2.4K]

Answer:

r= (-15) is the answer to the question ok

8 0
3 years ago
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Multiply. What is the value of the second partial product 27x108
jeyben [28]
<h2>Answer:</h2><h2>The  partial products of (27) (108) are 216 , 0 , 27.</h2>

Step-by-step explanation:

To find the value of the second partial product 27x108,

Here the multiplicand = 27

The multiplier = 108

The partial products of the multiplication are :

(27) (8)   =  0 2 1 6

(27) (0)   =  0 0 0

(27) (1)    =  2 7

Adding the terms, we get 2916.

The partial products are 216 , 0 , 27.

5 0
3 years ago
Pls help me on this math watch!!! this is non calculator
Andru [333]

The height of the cone is 4 cm (approx).

Step-by-step explanation:

Given,

The radius (r) of the cone = 4.96 cm

Volume (V) = 102 cm³

To find the height (h) of the cone.

Formula

The volume of a cone V = \frac{1}{3}πr²h

Now,

According to the problem,

\frac{1}{3}πr²h = 102

or,  \frac{1}{3}×\frac{22}{7}×(4.96)²h= 102

or, h = \frac{102X7X3}{4.96^{2}X22 }

or, h = 3.95 = 4 (approx)

so,

Height = 4 cm (approx)

8 0
3 years ago
PLEASE HELPPP I HAVE 2 HOURS TO GET DONE 15 ASSIGNMENTS PLUS A FINAL EXAM, WILL GIVE BRAINY TO CORRECT ANSWER
Katarina [22]

Answer:

13, 22, 37, 61

Step-by-step explanation:

7 0
3 years ago
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