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3 years ago

(3x^2y^6)^7 please help me

Mathematics

1 answer:

DerKrebs [107]3 years ago

5 0

Answer:

I'm not sure but I think the answer is 2187 X^ (14) Y^ ( 42)

Step-by-step explanation:

(3x^2y^6)^7

3^(7) x (X^)^(7) x ( Y ^(6)^(7)

= 2187 X^ 14)Y^( 42)

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Suppose you just purchased a digital music player and have put 12 tracks on it. After listening to them you decide that you like

Vilka [71]

Answer:

a) 9% , not unusual

b) 42.4%

c) 48.4%

d) 11.1% , 44.4% , 44.4%

Step-by-step explanation:

We have the following information from the statement:

n = 12

r = 4

P (likebothofthem) = P (likefirstsong) * P (likesecondsong)

P = 4/12 * 3/11

P = 0.09 = 9%

The probability is not unusual, unusual is considered less than 0.05 or 5%

P (likeneither) = P (notlikefirstsong) * P (notlikesecondsong)

P = 8/12 * 7/11

P = 0.424 = 42.4%

c) P (likeexactlyoneofthem) = P (firstsongliked) * P (secondsongnotliked) + P (firstsongnotliked) * P (secondsongliked)

P = (4/12 * 8/11) + (8/12 * 4/11)

P = 0.484 = 48.4%

P (likebothofthem) = P (likefirstsong) * P (likesecondsong)

P = 4/12 * 4/12

P = 0.111 = 11.1%

The probability is not unusual, unusual is considered less than 0.05 or 5%

P (likeneither) = P (notlikefirstsong) * P (notlikesecondsong)

P = 8/12 * 8/12

P = 0.444 = 44.4%

c) P (likeexactlyoneofthem) = P (firstsongliked) * P (secondsongnotliked) + P (firstsongnotliked) * P (secondsongliked)

P = (4/12 * 8/12) + (8/12 * 4/12)

P = 0.444 = 44.4%

8 0

3 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

3 years ago

Solve the following equation (If possible please show work)

irina [24]

Answer:

Step-by-step explanation:

5(x + 4) - 8 = x + 32

5x + 20 - 8 = x + 32

5x + 12 = x + 32

5x - x + 12 = 32

5x - x = 32 - 12

4x = 32 - 12

4x = 20

x = 5

Hope this helped! :)

4 0

3 years ago

Select the correct answer.

ivanzaharov [21]

The weather that is most likely to occur where Richard is is rain shower.

<h3>What type of weather is most likely to occur?</h3>

Weather is the state of the atmosphere of a particular place over a short period of time. There is low atmospheric pressure, so it is likely that it would rain. This is because when the atmospheric pressure is low, it is likely that it would rain.

To learn more about weather, please check: brainly.com/question/9382618

#SPJ1

7 0

2 years ago

Read 2 more answers

Need ASAP Now!! please help with answer!

puteri [66]

Answer:

$y = 1500 * 1.09^t$

Step-by-step explanation:

$\text{It should be clear that the rate of change is non-linear}$ $\text{, meaning that we can express the function in the following manner}$

$\text{y = a * b^t, where a is the initial value and b is the rate of change and t is how}$ $y = a * b^t,$ $\text{where a is the initial value and b is the rate of change and t is how many years.}$