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3 years ago

Darryl is considering a purchase of a $125,000 home at 6.5%. Under this rate,

Mathematics

1 answer:

raketka [301]3 years ago

3 0

Answer:

122 months or just over 10 years

Step-by-step explanation:

You might be interested in

Simplify 6(x+2)-(4x-14) the sum of two conicuative integers is -67. find the integer

olga_2 [115]

$\tt \: 6(x + 2) - (4x - 14)$

Distribute 6 through the parentheses

$\tt6x + 12 - 4(4x - 14)$

Distribute -4 through the parentheses

$\tt6x + 12 - 16x + 56$

Collect like terms

$\tt - 10x + 68$

$\rule{300pt}{3pt}$

Let the two integers be x and x+2

$\sf \: x + x + 2 = - 67$

Let's solve this !~

$\sf \: 2x + 2 = - 67$

$\sf \: 2x = - 67 - 2$

$\sf \: 2x = - 69$

$\sf \: x = - \frac{69}{2}$

$\sf \: x = - 34.5$

$\rule{300pt}{3pt}$

$\large{|\underline{\mathtt{\red{N}\blue{u}\orange{m}\pink{b}\blue{e}\purple{r}\green{\:}\red{1}\blue{:}\orange{-}}}}$

$\sf = \: x \\ \sf \: = - 34.5$

$\large{|\underline{\mathtt{\red{N}\blue{u}\orange{m}\pink{b}\blue{e}\purple{r}\green{\:}\red{2}\blue{:}\orange{-}}}}$

$\sf \: = x + 2 \\ \sf = - 34.5 + 2 \\ \sf \: = - 32.5$

8 0

2 years ago

Suppose that Y has density function

zvonat [6]

I'm assuming

$f(y)=\begin{cases}ky(1-y)&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(a) f(x) is a valid probability density function if its integral over the support is 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1 y(1-y)\,\mathrm dy=k\int_0^1(y-y^2)\,\mathrm dy=1$

Compute the integral:

$\displaystyle\int_0^1(y-y^2)\,\mathrm dy=\left(\frac{y^2}2-\frac{y^3}3\right)\bigg|_0^1=\frac12-\frac13=\frac16$

So we have

k / 6 = 1 → k = 6

(b) By definition of conditional probability,

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4 and Y ≤ 0.8) / P(Y ≤ 0.8)

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4) / P(Y ≤ 0.8)

It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since F(y) = P(Y ≤ y).

We have

$\displaystyle F(y)=\int_{-\infty}^y f(t)\,\mathrm dt=\int_0^y6t(1-t)\,\mathrm dt=\begin{cases}0&\text{for }y$

Then

P(Y ≤ 0.4) = F (0.4) = 0.352

P(Y ≤ 0.8) = F (0.8) = 0.896

and so

P(Y ≤ 0.4 | Y ≤ 0.8) = 0.352 / 0.896 ≈ 0.393

(c) The 0.95 quantile is the value φ such that

P(Y ≤ φ) = 0.95

In terms of the integral definition of the CDF, we have solve for φ such that

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=0.95$

We have

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=\int_0^\varphi 6y(1-y)\,\mathrm dy=(3y^2-2y^3)\bigg|_0^\varphi = 0.95$

which reduces to the cubic

3φ² - 2φ³ = 0.95

Use a calculator to solve this and find that φ ≈ 0.865.

8 0

3 years ago

a catering service offers 6 appetizers, 10 main courses, and 8 desserts. A banquet committee is to select 2 appetizers, 3 main c

S_A_V [24]

There are two different answers that you could be looking for.
You might be asking how many different meals can be served at the banquet,
or you might be asking literally how many 'ways' there are to put meals together.

I'm going to answer both questions. Here's how to understand the difference:

Say you have ten stones, and you tell me "I'll let you pick out two stones
and take them home. How many ways can this be done ?"

For my first choice, I can pick any one of 10 stones. For each of those . . .
I can pick any one of the 9 remaining stones for my second choice.
So the total number of 'ways' to pick out two stones is (10 x 9) = 90 ways.

But let's look at 2 of those ways:
 -- If I pick stone-A first and then pick stone-G, I go home with 'A' and 'G'.
 -- If I pick stone-G first and then pick stone-A, I still go home with 'A' and 'G'.
There are two possible ways to pick the same pair.
In fact, there are two possible ways to pick every pair.
So there are 90 ways to pick a pair, but only 45 different pairs.

That's the reason for the difference between the number of ways the
committee can make their selections, and the number of different meals
they can put together for the banquet.

So now here's the answer to the question:

-- Two appetizers can be selected in (6 x 5) = 30 ways.
(But each pair can be selected in 2 of those ways,
so there are only 15 possible different pairs.)

-- Three main courses can be selected in (10 x 9 x 8) = 720 ways.
(But each trio can be selected in 3*2=6 of those ways,
so there are only 120 possible different trios.)

-- Two desserts can be selected in (8 x 7) = 56 ways.
(But each pair of them can be selected in 2 of those ways,
so there are only 28 possible different pairs.)

-- The whole line-up can be selected in (30 x 720 x 56) = 1,209,600 ways.

But the number of different meals will be (30 x 720 x 56) / (2 x 6 x 2) =

 (15 x 120 x 28) = 50,400 meals.

5 0

3 years ago

You purchased 8 pounds 10 ounces of candy shop you want to split it equally among 3 classrooms at a local school. How much shoul

amid [387]

Given than 1 pound = 16 ounces, we have a total of 8*16 + 10 = 138 ounces which we must devide among the 3 classrooms. This gives us 138/3 = 46 ounces per classroom, or 2 pounds and 14 ounces per classroom.

8 0

3 years ago

In the past, the mean running time for a certain type of flashlight has been 9.2 hours. The manufacturer has introduced a change

ELEN [110]

Answer:

what are the hypotheses?

Step-by-step explanation:

you said the hypotheses are as folows its not there

6 0

3 years ago