Guys I am giving you a list of what i need help with i did the rest

Hello, I'm new! I need help

melamori03 [73]

The interest in the first month is given as $ 97.1. The principal balance in the second question is $15,030.02

<h3>How to solve for the interest in the first month</h3>

1. We have to solve for the cost of the car

This would be = 19,725*(1.0475)

= 20,661.9375

There is a Down payment = 2,175

balance would be 20661.9375-2,175 = 18,486.94

average rating interest of new car = 6.30%

So the interest accrued in first month = 18,486.94x0.063/12 = $ 97.1

2. cost = 15867

sales tax = 5.25%

10 percent down payment

5.25/100 = 0.0525

cost of car = 15867 + (15867 * 0.0525)

= 16700 dollars

10% of 16700 dollars

= 1670 dollars

principal balance = 16700 - 1670

= $15,030.02

Read more on interest rate here

brainly.com/question/25793394

7 0

1 year ago

Segment AB has endpoints with coordinates A: (-4, -7) and B: (8, -4). Determine the value of the x-coordinate of the interior po

Nadusha1986 [10]

Answer:

≈ <em>3.4</em><em> </em>

Step-by-step explanation:

Coordinates of the point that divide segment with endpoints A( $x_{1}$ , $y_{1}$ ) and B( $x_{2}$ , $y_{2}$ ) in a ratio a : b are

$\frac{bx_{1} +ax_{2} }{a+b}$ , $\frac{by_{1} +ay_{2} }{a+b}$ )

A(- 4, - 7)

B(8, - 4)

Ratio 8 : 5

x-coordinate is $\frac{5(-4)+8(8)}{8+5}$ = $\frac{44}{13}$ ≈ <em>3.4</em>

3 0

2 years ago

What is the measure of Y? 4·9=36=36/4

alexira [117]

Answer:

Figure or Diagram Mandatory

GOOD LUCK FOR THE FUTURE! :)

3 0

2 years ago

I will mark you as brainliest if you're answer is correct

GalinKa [24]

Answer:

perimeter = 28.13 cm

Step-by-step explanation:

Calculate the length of a line drawn from A to C.

sin 85 = b/8, b = 7.97

cos 85 = c/8, c = 0.697

AC² =7.97² + (6 + 0.697)² = 63.52 + 44.85 = 108.37

AC = 10.41

Now you have a right triangle ADC with a known hypotenuse and one leg.

Using the Pythagorean theorem:

DC² = 10.41² - 5²

DC² = 108.37 - 25 = 83.37

DC = 9.13 cm

perimeter = 5 + 6 + 8 + 9.13 = 28.13 cm

8 0

2 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago