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borishaifa [10]
3 years ago
8

\frac{3-\frac{a}{b} }{\frac{1}{b}+b }[/tex]

Mathematics
1 answer:
hjlf3 years ago
6 0

Hello from MrBillDoesMath!

Answer:

(3b -a) / ( 1 + b^2)

Discussion:

My best guess is that we are trying to simply this fraction:

( 3 - a/b) / ( (1/b) + b)   =             => multiply num. and denom. by "b"

(3b - a) / (   (1/b)*b + b*b) =

(3b -a) / ( 1 + b^2)


Thank you,

MrB

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Please help !!!!!!!!!!!!
zhannawk [14.2K]

Answer:

25+40

Step-by-step explanation:

for the area of a triangle you multiple the base times the height. for the area of the square it is the length times width.

NOTE: to completely answer just add 25+40

6 0
3 years ago
How do i plot this??
Softa [21]

Answer:

its pretty simple actually put a point after the -1 (-0.9) and move left twice from -1 to get the plot point -1.2

5 0
3 years ago
A=4pi r^{2} solve for r
lara [203]
             A = 4πr²          Divide both sides by 4π
     A / 4π = r²               Find the square root of both sides
√(A / 4π) = √(r²)          Cancel out the square with the square root
√(A / 4π) = r                Switch the sides to make it easier to read
              r = √(A / 4π)
5 0
2 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

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

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
3 years ago
wind resistance varies jointly as an objects surface area and velocity. if an object traveling at 55 miles per hour with a surfa
nikdorinn [45]

Answer:

25 miles per hour

Step-by-step explanation:

Given

W = Wind\ resistance

A = Surface\ Area

V = Velocity

The joint variation can be represented as:

W\ \alpha\ A*V

Where:

V = 55; A = 20; W = 220

Required

Find V,  when: A = 55; W = 275

We have:

W\ \alpha\ A*V

Express as an equation

W= k *A*V

Where k is the constant of variation

Make k the subject

k = \frac{W}{A*V}

When: V = 55; A = 20; W = 220

We have:

k = \frac{220}{20 *55 }

k = \frac{220}{1100}

k = 0.2

When: A = 55; W = 275

We have:

k = \frac{W}{A*V}

Substitute: A = 55; W = 275 and k = 0.2

0.2 = \frac{275}{55 * V}

Make V the subject

V= \frac{275}{55 * 0.2}

V= \frac{275}{11}

V= 25

3 0
3 years ago
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