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fenix001 [56]
2 years ago
8

Can someone pls help

Mathematics
1 answer:
enyata [817]2 years ago
5 0

Answer:

b

Step-by-step explanation:

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Solve the following equation: 156 + (7 – 3) x 7 = A. 174 B. 184 C. 1,050 D. 1,120
IrinaVladis [17]
In this problem, it is essential to use the order of operations, or PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction). First, you solve 7-3=4. Then, you multiply 7 times 4, which is 28. Lastly, you add 28 to 156 which is 184. So your answer is B.
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What is the positive root of the equation x2 + 5x = 150?
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Answer:

10

Step-by-step explanation:

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The diagram shows the measurements of a frame into which concrete is poured to make steps. The steps are shaped like rectangular
Ivahew [28]
(6 x 16 x 6) + (14 x 16 x 7)
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3 years ago
15. Write the slope-intercept form of the line described in the following:Parallel to 4x + 5y=20and passing through (12,4)
Amiraneli [1.4K]

Problem

Write the slope-intercept form of the line described in the following:

Parallel to 4x + 5y=20

and passing through (12,4)

Solution>

For this case we need to have the same slope, and if we write the equation given we see:

5y = 20 -4x

y = 4 -4/5 x

then the slope m = -4/5

and we also know a point given x= 12, y= 4 and we can do the following:

4 = -4/5 (12) +b

4 = -48/5 + b

And if we solve for the intercept we got:

b= 4 +48/5= -28/5

And our equation would be given by:

y = -4/5 x -28/5

3 0
1 year ago
Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.
Sindrei [870]

Answer:

a) P(X=3) = 0.1

b) P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

c) P(X=4) = 0.3

d) P(X=0) = 0.2

e) E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

f) E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

P(X=3) = 0.1

Part b

We want this probability:

P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

Part c

For this case we want this probability:

P(X=4) = 0.3

Part d

P(X=0) = 0.2

Part e

We can find the mean with this formula:

E(X)= \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

Part f

We can find the second moment with this formula

E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

4 0
3 years ago
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