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siniylev [52]
3 years ago
7

Tahmid has 96 G.I. Joe action figures and Sandeep has one-third as much as Tahmid does. How many figures must Tahmid give Sandee

p in order for them to have the same number of figures?
Mathematics
1 answer:
Ann [662]3 years ago
6 0
Tahmid has 96 figures, Sandeep has 1/3 of it so 
96(1/3) = 96/3 = 32
Sandeep has 32 figures

Tahmid 96 ; Sandeep 32

How many must tahmid give so they have the same figures?
First take the difference of the two so we know how many tahmid can give. 
96-32 = 64
now they each have 32 and 64 give or keep. for each one tahmid gives, he must keep one himself so to match the number they both own. so we divide the number of 2, one part to give, the other part to keep himself.
64/2 = 32
So, Tahmid must give Sandeep 32 figures to have equal number of figures of 64 each. 
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Helga [31]

Answer:

(a)96.77%

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Step-by-step explanation:

Starting with the Michaelis-Menten equation which is used to model biochemical reactions:

Dividing both sides by V_{\max }}

\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}

Where: V_{\max }} = maximum rate achieved by the system

K_{\mathrm {M} }}=The Michaelis constant

{\displaystyle {\ce {[S]}}}= Substrate concentration

(a) When [S]=30K_M

\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}\\\dfrac{v}{V_{max}}=\dfrac{30K_M}{K_M + 30K_M}\\\dfrac{v}{V_{max}}=\dfrac{30}{1 + 30}\\\dfrac{v}{V_{max}}=\dfrac{30}{31}\\$Expressed as a percentage\\\dfrac{v}{V_{max}}=\dfrac{30}{31}X100=96.77\%

(b)When K_M=30[S]

\dfrac{v}{V_{max}}=\dfrac{[S]}{K_M + [S]}\\\dfrac{v}{V_{max}}=\dfrac{[S]}{30[S] + [S]}\\\\=\dfrac{1[S]}{30[S] + 1[S]}\\=\dfrac{1}{30 + 1}\\\dfrac{v}{V_{max}}=\dfrac{1}{31}\\$Expressed as a percentage\\\dfrac{v}{V_{max}}=\dfrac{1}{31}X100=3.23\%

8 0
3 years ago
Round 7 3/8 to the nearest whole number?!?
DIA [1.3K]

Answer:

7

Step-by-step explanation:

Everything including 1/2 and above rounds up to the next whole number.

3/8 < 4/8

7 and 3/8 rounds to 7

5 0
2 years ago
Evaluate the line integral, where c is the given curve. c y3 ds,
postnew [5]
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\mathbf r(t)=(t^3,t)

with 0\le t\le2, we have

\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt

So

\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt
=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du
=\dfrac{145^{3/2}-1}{54}
5 0
3 years ago
A coin is tossed 4 times. Let E1 be the event "the first toss shows heads" and E2 the event "the second toss shows heads" and so
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Answer:

The events are independent.

The probability of showing heads on both toss is equal to 1/2

Step-by-step explanation:

The sample space for this experiment consists of 2⁴= 16 sample points, as each toss can result in two outcomes we assume that the events are equally likely.

Two events are independent in the sample space if the probability of one event occurs, is not affected by whether the other event has or has not occurred.

In general the k events are defined to be mutually independent if and only if the probability of the intersection of  any 2,3,--------, k  equals the product of their respective probabilities.

P (A∩B) = P(A). P(B)

P (A∩B)   = 1/2. 1/2= 1/4

                                                                  Head          Tail

 P(E1)= 1/2  ----------          Coin 1               H,H              T,H

                                                                1/4                  1/4

  P(E2)= 1/2  ---------------  Coin 2             H, H               H,T

                                                                      1/4           1/4

So the events are independent.

The probability of showing heads on both toss is equal to 1/2

The sample space for this experiment consists of 2⁴= 16 sample points, out of which eight will have heads on both toss.

Or in other words ( 1/4* 1/4) = 2/4 = 1/2

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