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3 years ago

Which ordered pair is a solution to the system of inequalities graphed here?

Mathematics

1 answer:

saul85 [17]3 years ago

6 0

Answer: I dont know

Step-by-step explanation:

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What is 3/5 times 6/5

Angelina_Jolie [31]

Answer is 0:72

hope it is helpful

6 0

3 years ago

Read 2 more answers

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

How do you find how many times larger

masya89 [10]

Answer

whats the problem.

Step-by-step explanation:

4 0

3 years ago

Solve the variable in the following proportion. 3.6/y=1.2/2

Molodets [167]

Hello :
3.6/y=1.2/2
1.2 y = 2(3.6)
y = 2(3.6)/1.2
y=6

7 0

3 years ago

Assume that the weight of two year old babies have distribution that is approximately normal with a mean of 29 pounds and a stan

Lana71 [14]

Answer:

25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 29 pounds

Standard Deviation, σ = 3 pounds

We are given that the distribution of weight of two year old babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.10

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 29}{3})=0.10$

Calculation the value from standard normal z table, we have,

$P(z < -1.282) = 0.10$

$\displaystyle\dfrac{x - 29}{3} = -1.282\\x = 25.154 \approx 25.15$

Thus, 25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

5 0

3 years ago