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LenaWriter [7]
3 years ago
6

A) Show work, b) solve, and c) graph the solutions for −7+2x>−27

Mathematics
2 answers:
8_murik_8 [283]3 years ago
7 0

Answer:

x > -10

Step-by-step explanation:

−7+2x>−27

Add 7 to each side

−7+7+2x>−27+7

2x > -20

Divide each side by 2

2x/2 > -20/2

x > -10

Zanzabum3 years ago
6 0

Answer:

Solve:

-7 + 2x > -27

Add both sides by 7:

-7 + 7 +2 > -27 +7

2x > -20

-Divide both sides by 2:

\frac{2x}{2} > \frac{-20}{2}

x > -10

After you have the answer x > -10, circle the mark above -10 on the number and draw the arrow to the right.

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Estimate the Value Of Each Expression to the nearest integer .
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see below

Step-by-step explanation:

14. We can represent 1 as √1 and 2 as √4 and because 1 < 3 < 4, we know that √3 is in between 1 and 2. However, 3 is closer to 4 than it is to 1 so √3 is about 2.

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16. -4 = -√16 and -5 = -√25, since 16 < 22 < 25, we know that -√22 is in between -4 and -5 but 22 is closer to 25 so the answer is -5.

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5 0
3 years ago
It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

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In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

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Finally, we have that the probability of returning exactly one of the three packages is 1.27%

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Answer:

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Step-by-step explanation:

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Hope this helps!

6 0
4 years ago
Read 2 more answers
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