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taurus [48]
4 years ago
14

Solve the following using scientific notation and standard notation:

Mathematics
2 answers:
victus00 [196]4 years ago
4 0
<span>1) 2.5 x 10^7 
= 25,000,000

2) 1 x 10^-3 
= 0.001

3) 3,670,000
= 3.67 x 10^6

4) .00000000295
=2.95 x 10^-9</span>
Setler79 [48]4 years ago
3 0
1: 25000000
2: 1/1000

and the other 2 i don't understand
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If the probability of being hospitalized during a year is 0.15 find the probability that no one in the family of 4 will be hospi
Goryan [66]

The probability that no one in the family of 4 will be hospitalized during a year is 0.52

Step-by-step explanation:

The probability for a person to be hospitalized during one year is

p(h)=0.15

As a consequence, the probability for a person of NOT being hospitalized during one year is

p(h^{-1})=1-0.15=0.85

If we have 4 people in the family, the probability of each of them of NOT being hospitalized during the year is independent from the probability of the others - therefore, these are independent events. This means that the total probability of the 4 people of not being hospitalized during a year is:

p(h^{-1})^4=(0.85)^4=0.52

So, the probability that no one in the family of 4 will be hospitalized during a year is 0.52.

Learn more about probability:

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3 0
3 years ago
Jake earns $7.50 per hour working at a local car wash. The function, ƒ(x) = 7.50x, relates the amount Jake earns to the number o
Pie
F -1 (x) = x/7.50

Hope this helps!
6 0
3 years ago
I need help in partial fraction!! With simple explaination would be nice!
WITCHER [35]
\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}

Now the second term can be decomposed into partial fractions.

\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}
\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}
4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)
4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1
\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3
\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}

So

\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}

3 0
3 years ago
Terrence finished a word search in 3/4 the time it took frank. charlotte finished the word search in 2/3 the time it took terren
Luden [163]

Answer:

Charlotte-16

Step-by-step explanation:

In order to calculate the time that Charlotte took we just first have to calculate the time that Terrence took to finish the word search, that is done by multiplying the 32 minutes that Frank took by the three quartes that it took terrence:

Time=(\frac{3}{4}) (32)\\Time=(\frac{96}{4})\\Time=24

So now we know that It took terrence 24 minutes, and that Charlotte took two thirds of that time so we now just multiply that:

Time=(\frac{2}{3}) (24)\\Time=(\frac{48}{3})\\Time=16

6 0
4 years ago
Read 2 more answers
How do I find the apothem of a decagon with the radius of 14???
Bogdan [553]

Answer:

A.  13.31'  to the nearest hundredth.

B.  86.52' to the nearest hundredth.

Step-by-step explanation:

Part A>

The apotherm of a regular polygon is the distance of the line segment from the centre of the polygon to the midpoint of a side.

2 radii of this decagon joined to the endpoints of a side form an isosceles triangle with equal sides = 14 cm.

The apotherm is the altitude of this triangle.  The  vertex of the triangle  has an angle of 360 / 10 = 36 degrees and the apotherm bisects this angle.

So using trigonometry on the right triangle formed:

cos 18 = x / 14   where x is the apotherm.

x = 14 cos 18

=  13.31'  (answer).

Part B.

Using trigonometry on the right triangle again:

sin 18 = x/2 / 14     (where x is the length of a side of the hexagon)

x/2 = 14 sin 18

x = 2 * 14 sin 18

= 8.652'.

As a decagon has 10 sides the perimeter = 8.652 * 10

= 86.52' to the nearest hundredth.

6 0
3 years ago
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