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Alenkinab [10]
3 years ago
5

What is the decimal expansion of the following fraction? 1/8 A. 0.125 B. 0.125 C. 1.8 D. 0.18​

Mathematics
1 answer:
qaws [65]3 years ago
3 0

Answer:

0.125....................

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13. A new bank customer with $5,000 wants to open a money market account. The bank is
9966 [12]

Answer:

a) 900 dollars

b) 5900 dollars

Step-by-step explanation:

The complete question is

A new bank customer with $5,000 wants to open a money market account. The bank is offering a simple interest rate of 1.8%. a. How much interest will the customer earn in 10 years? b. What will the account balance be after 10 years?

7 0
3 years ago
Luke subtracted 128° from 180° and found that the measurement of ∠STR is 52°. How can Luke use this information to find the meas
Karolina [17]
STR is a triangle as shown in the picture. If Luke decided to subtract 128º from 180º, that's because he understood that the sum of the internal angles in a triangle is 180º, and therefore the angles RST and TRS together make 128º.
As a result, RST will be
128º - TRS
where TRS is the angle at R, as shown in the picture

6 0
3 years ago
Read 2 more answers
y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence
sukhopar [10]

Let

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots

Differentiating twice gives

\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots

\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0

\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0

Then the coefficients in the power series solution are governed by the recurrence relation,

\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

k=0 \implies n=0 \implies a_0 = a_0

k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}

k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}

k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

k = 0 \implies n=1 \implies a_1 = a_1

k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}

k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}

k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}

so that

a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}

So, the overall series solution is

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)

\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

4 0
3 years ago
Find the compunt intrest on rs 31250 at 12 % per annum for 12 1/2 years
postnew [5]

Answer:

<h2>$128847</h2>

Step-by-step explanation:

given Principle at 31250

and rate = 12%

and time 12.5 years

the amount would be given by the equation

P(1+\frac{rate}{100})^{time}

plug in values to get

31250(1+\frac{12}{100})^{12.5}

end up with value $128847

3 0
2 years ago
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
VMariaS [17]

Answer:

He must survey 123 adults.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

Assume that a recent survey suggests that about 87​% of adults have heard of the brand.

This means that \pi = 0.87

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How many adults must he survey in order to be 90​% confident that his estimate is within five percentage points of the true population​ percentage?

This is n for which M = 0.05. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.05 = 1.645\sqrt{\frac{0.87*0.13}{n}}

0.05\sqrt{n} = 1.645\sqrt{0.87*0.13}

\sqrt{n} = \frac{1.645\sqrt{0.87*0.13}}{0.05}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.87*0.13}}{0.05})^2

n = 122.4

Rounding up:

He must survey 123 adults.

6 0
3 years ago
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