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3 years ago

20 This spinner will be spun twice. What is the probability of

Mathematics

1 answer:

natali 33 [55]3 years ago

3 0

Answer: the probability is 0.25

Step-by-step explanation:

We have 10 numbers:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Of those, the odd ones are:

1, 3, 5, 7, 9

So we have 5 odd numbers.

The probability that the outcome is an odd number is equal to the number of odd numbers divided the total number of numbers:

p = 5/10 = 0.5

For the second spin the probability is the same, p = 5/10, because the first outcome does not affect the results of the second spin.

The probability of spining an odd number both times, then is the joint probability for two times this same event:

P = (5/10)(5/10) = 0.5*0.5 = 0.25

or 25% in percent form

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Simplify -3d^8(-4d^-14). Assume d =/ 0

Alenkinab [10]

Answer:

$\frac{12}{d^6}$

Step-by-step explanation:

$-3d^8\left(-4d^{-14}\right)$

Double negative = Positive:

$3d^8*4d^{-14}$

Apply negative exponent formula:

$3*d^8*4*\frac{1}{d^{14}}$

Multiply:

$\frac{12d^8}{d^{14}}$

Apply exponent rule: $\frac{x^a}{x^b}=\frac{1}{x^{b-a}}$

Thus, $\frac{12}{d^{14-8}}$

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$\frac{12}{d^6}$

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2 years ago

Elena has $0.90. Kia has 10 times as much money as Elena. Troy has one-tenth as much money as Elena. Does Kia have more money th

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2 years ago

Read 2 more answers

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3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

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3 years ago