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Ilya [14]
3 years ago
9

Assume each newborn baby had a probability of approximately 0.54 of being female and 0.46 of being male binomial distribution?

Mathematics
1 answer:
Dmitry [639]3 years ago
4 0
0.54 is the binomial so yea 46
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Please answer
SVEN [57.7K]

From the calculation done below, it can be seen that the cost of one orange is 85 cents.

<h3>How do we determine the cost of a product?</h3>

The cost of one orange can be determined as follows:

One dime = 10 cents

2 dollars = 200 cents

Total amount given by Sue to the store clerk = One dime + 2 dollars = 10 cents + 200 cents = 210 cents

Cost of two oranges = Total amount given by Sue to the store clerk –

Change = 210 cents – 40 cents = 170 cents

Cost of one orange = Cost of two oranges / 2 = 170 cents / 2 = 85 cents

Learn more about cost here: brainly.com/question/14844899.

#SPJ1

7 0
2 years ago
Dose anyone know how to do this
Neko [114]
The first one is < the second one is < and the last one is =
5 0
4 years ago
Read 2 more answers
What is f(-3) if f(x) =|2x-1|-5
dangina [55]
F(x) = |2x-1|-5

f(x) = |2*(x)-1|-5

f(-3) = |2*(-3)-1|-5 ... replace every x with -3

f(-3) = |-6-1|-5

f(-3) = |-7|-5

f(-3) = 7-5

f(-3) = 2 which is the final answer

Side Note: the vertical bars mean "absolute value" which represents distance. Negative distance makes no sense which is why absolute value results are never negative.

6 0
3 years ago
Read 2 more answers
Francesca is afraid she won't have enough money to treat herself and two friends to a movie and snacks. Movie tickets cost $7.50
Archy [21]
The correct answer is C.
8 + 8 + 8 + 3 + 3 + 3 + 3 + 3 + 3 = 42
7 0
3 years ago
Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant be
Leona [35]

In polar coordinates, the region R is the set of points

\left\{(r,\theta)\mid3\le r\le5,\,0\le\theta\le\dfrac\pi2\right\}

and we have x^2+y^2=r^2 and \mathrm dA=r\,\mathrm dr\,\mathrm d\theta. So the integral, converted to polar, is

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_3^5r\sin r^2\,\mathrm dr\,\mathrm d\theta=\frac\pi2\int_3^5r\sin r^2\,\mathrm dr

Substitute s=r^2 to get

=\displaystyle\frac\pi4\int_9^{25}\sin s\,\mathrm ds=\frac{(\cos9-\cos25)\pi}4

5 0
3 years ago
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