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mario62 [17]
3 years ago
6

Please help me with this. ​

Mathematics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

a) 48 b) 132 c) 180 d) 228 e) 312

Step-by-step explanation:

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Which ordered pair is a solution of the equation?
True [87]

Answer:

Step-by-step explanation: try your best

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2 years ago
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Drag each scenario to show whether the final result will be greater than the original
Hoochie [10]

A $20 decrease followed by a $20 increase - original value

A 75% decrease followed by a 50% increase - less than the original value

A 50% increase followed by a 33 1/3% decrease - greater than the original value

A 60% increase followed by a 40% decrease - greater than the original value

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3 years ago
3-31÷4=2<br><br> 15points if you answer
Marat540 [252]

Step-by-step explanation:

-28÷4=2

-7=2

= -7-2

= -9 is the correct answer..

6 0
2 years ago
What is 6/a-6 = 3/a-3 ?<br> help me please
Alenkasestr [34]

Answer:

a=1

Step-by-step explanation:

Multiply all terms by a and cancel:

6+−6a=3+−3a

−6a+6=−3a+3(Simplify both sides of the equation)

−6a+6+3a=−3a+3+3a(Add 3a to both sides)

−3a+6=3

−3a+6−6=3−6(Subtract 6 from both sides)

−3a=−3

−3a

−3

=

−3

−3

(Divide both sides by -3)

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3 0
3 years ago
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The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in
Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 64 inches

           \sigma = standard deviation = 4 inches

           n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{63-64}{\frac{4}{\sqrt{4} } } ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

                                                        = 1 - 0.69146 = 0.30854

Hence, it is 30.85%  likely that average height would be shorter than 63 inches.

7 0
3 years ago
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