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3 years ago

12

Write an expression that represents the number of videos Charli will have in w weeks.

1 answer:

sweet [91]3 years ago

5 0

Ummmmmmmmmmmmmmmmmmm

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10% of a group is left handed. If a random sample of 12 switches is selected, what is the probability of exactly two of the swit

Paraphin [41]

Assuming that 10% is the probability of a switch being defective (since left-handedness has nothing to do with the problem):
For a binomial distribution, probability(r out of n) = (nCr) (p)^r (q)^(1-r)
p = 10% = 0.1, q = 1 - p = 0.9
r = 2 switches, n = total of 12 switches
probability = (12C2) (0.1)^2 (0.9)^(12-2)
probability = 66(0.1)^2 (0.9)^10
probability = 0.23

4 0

3 years ago

Expand the expression.<br> 0.2(y + 2)

Stolb23 [73]

Answer:

x=0.2(y+2)

Step-by-step explanation:

7 0

3 years ago

A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount

Alja [10]

Answer:

$x = 97$

Step-by-step explanation:

Given

$t = 20$ --- time (years)

$A =1000$ --- amount

$r = 10\%$ --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

$y_1(end) = 10\% * 1000 * 150\%$

$y_1(end) = 150$

Amount at end of year 1

$A_1=A - y_1(end) - r * A$

$A_1=1000 - (150 - 10\% * 1000)$

$A_1 =1000 - (150- 100)$

$A_1 =950$

Rewrite as:

$A_1 = 0.95 * 1000^1$

Next, calculate the end of year 1 payment

$y_2(end) = 10\% * 950 * 150\%$

$y_2(end) = 142.5$

Amount at end of year 2

$A_2=A_1 - (y_2(end) - r * A_1)$

$A_2=950 - (142.5 - 10\%*950)$

$A_2 = 902.5$

Rewrite as:

$A_2 = 0.95 * 1000^2$

We have been able to create a pattern:

$A_1 = 1000 * 0.95^1 = 950$

$A_2 = 1000 * 0.95^2 = 902.5$

So, the payment till the end of the 10th year is:

$A_{10} = 1000*0.95^{10}$

$A_{10} = 598.74$

To calculate X (the last 10 payments), we make use of the following geometric series:

$Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n$

$Amount = \sum\limits^{9}_{n=0} x * (1.10)^n$

The amount to be paid is:

$Amount = A_{10}*(1 + r)^{10}$ --- i.e. amount at the end of the 10th year * rate of 10 years

$Amount = 1000 * 0.95^{10} * (1+r)^{10}$

So, we have:

$Amount = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}$

$\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}$

The geometric sum can be rewritten using the following formula:

$S_n = \sum\limits^{9}_{n=0} x * (1.10)^n$

$S_n =\frac{a(r^n - 1)}{r -1}$

In this case:

$a = x$

$r = 1.10$

$n =10$

So, we have:

$\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n$

$x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n$

So, the equation becomes:

$x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}$

Solve for x

$x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}$

$x = 97.44$

Approximate

$x = 97$

4 0

3 years ago

it takes a baseball about 9 seconds to come down from up in the air after a player hits homeroom suppose that the ball drops dow

Olenka [21]

Yes a is the correct answer

4 0

3 years ago

Daria is running a 30km race. She has completed 6km so far. What percent of the race has she completed?

Scorpion4ik [409]

Answer:

20%

Step-by-step explanation:

6/30 = .2 therefore she has ran 20% or 1/5 of the race

6 0

4 years ago