Which simplified question represents0.45

If Jessica purchases a property for $290,000, and borrows $200,000 she is said to have a LTV ratio of? 100%? 76%? 69%? 31%?

Luba_88 [7]

The loan-to-value ratio (LTV) is the ratio fo the first mortgage amount to the appraised value of the property. Here, that is
LTV = ($200,000)/($290,000)
LTV = 0.689655
LTV ≈ 69%

3 0

3 years ago

Read 2 more answers

If f(x)=3 x -1 and g(x)=0.03x^3-x+1 an approximate solution for the equation f(x)=g(x) is

lubasha [3.4K]

Answer:

x ≈ {-11.789, +0.501, 11.288}

Step-by-step explanation:

The cubic g(x) - f(x) = 0 has three real solutions. It can be rewritten as ...

$g(x)-f(x)=0\\\\0.03x^3-x+1-(3x-1)=0\\\\0.03x^3-4x+2=0$

Since the solutions are irrational, they are best found using a spreadsheet or graphing calculator. My favorite graphing calculator shows the approximate solutions below.

_____

Comment on the problem statement

Your expression for f(x) is ambiguous in that many Brainly questions have the exponentiation indicator replaced by a blank: 3 x -1 often means 3^x -1 and sometimes means 3^(x-1). We have taken the expression at face value and have assumed it is a linear expression. If otherwise, the problem is basically worked the same way: write a function h(x) = g(x) - f(x) and look for solutions to h(x) = 0. Graphing can be useful.

3 0

3 years ago

Historically, a certain region has experienced 65 thunder days annually. (A "thunder day" is day on which at least one instance

Makovka662 [10]

Answer:

No, we can't conclude that the mean number of thunder days is less than 65.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 65 thunder days annually. Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20.

We have to test that the mean number of thunder days is less than 65 or not.

Let, NULL HYPOTHESIS, $H_0$ : $\mu \geq$ 65 days {means that the mean number of thunder days is higher than or equal to 65}

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ < 65 days {means that the mean number of thunder days is less than 65}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of thunder days = 55

s = sample standard deviation = 20

n = number of years = 11

So, test statistics = $\frac{55-65}{\frac{20}{\sqrt{11} } }$ ~ $t_1_0$

= -1.658

Now, at 1% significance level t table gives critical value of -2.764. Since our test statistics is higher than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the left side of critical value rejection region.

Therefore, we conclude that the mean number of thunder days is higher than or equal to 65.

6 0

3 years ago

At 7:00 am, the temperature was -4℉. At 9:00 am, it was 8° warmer. What was the temperature at 9:00 am? *

Nana76 [90]

4℉.

What we know about Degrees is that there is a Positive type and a negative type.

(i.e: 30℉ is positive and -30℉ is negative.)

If the temperature was -4℉ at 7AM, then it is negative. If it goes up by an amount that is more than 4 then that negative will go up to a positive temperature. In this case: At 9AM it was 8° warmer.

Warmer is a keyword. If it is warmer by an amount, Negative temperatures will go up to a positive and positive temperature will just go up. If it gets cooler, negative temperatures will go down further and positive temperatures will go down to a negative.

So lets work out this problem with our newfound knowledge.

-4° F at 7AM

8° warmer at 9AM

-4 + 8 = 4.

The temperature was 4° at 9AM.

-Snooky

8 0

3 years ago

Stephen is 2 years older than Sushil. krutika is 4 years younger than Sushil. If the total of their ages is 61, how old is the e

german

Shushils age = 19.6
Stephens age 21.6
Krutikas age = 15.6