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3 years ago

Plz help me !!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Vlad1618 [11]3 years ago

5 0

It is A that is the answer your welcome

Sedaia [141]3 years ago

4 0

The answer is C lol

answer A is wrong, because it’s equal to 6b^2

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Quick computing company produces calculators they have found that the cost c(x),of making calculators is a quadratic function in

kirill115 [55]

A quadratic function is given by y = ax^2 + bx + c
c(2) = a(2)^2 + 2b + c = 45
4a + 2b + c = 45 . . . (1)
c(4) = a(4)^2 + 4b + c = 143
16a + 4b + c = 143 . . . (2)
c(10) = a(10)^2 + 10b + c = 869
100a + 10b + c = 869 . . . (3)

Solving (1), (2) and (3) gives a = 9, b = -5, c = 19

Therefore, c(x) = 9x^2 - 5x + 19
c(7) = 9(7)^2 - 5(7) + 19 = 9(49) - 35 + 19 = 441 - 16 = 425

Therefore, it costs $425 to produce 7 calculators.

4 0

3 years ago

Y=-2x+5<br> X | Y<br> ____<br> 5<br> -1<br> 2<br> 0

Y_Kistochka [10]

Answers:

Step-by-step explanation:

Plug in the x into the equation and solve.

<h3>Value 1: X = 5</h3>

Y = -2(5) + 5

Y = -10 + 5

Y = -5

<h3>Value 2: X = -1</h3>

Y = -2(-1) + 5

Y = 2 + 5

Y = 7

<h3>Value 3: X = 2</h3>

Y = -2(2) + 5

Y = -4 + 5

Y = 1

<h3>Value 4: X = 0</h3>

Y = -2(0) + 5

Y = 0 + 5

Y = 5

I'm always happy to help :)

8 0

3 years ago

The bad debt ratio for a financial institution is defined to the dollar value of loans defaulted divided by the total dollar val

vlabodo [156]

Answer:

Step-by-step explanation:

From the information given:

Consider X to be the random variable denoting the bad debt ratios for Ohio Bank.

Then, $X \sim N ( \mu, \sigma ^2)$

Thus the null hypothesis and the alternative can be computed as:

Null hypothesis:

$H_o : \mu \leq3.5\%$

Alternative hypothesis

$H_1 : \mu > 3.5\%$

The type I and type II error is as follows:

Type I:

The mean bad debt ratio is > 3.5% when it is not

Type II:

The mean bad debt ratio is ≤ 3.5% when it is not.

The test statistics can be calculated by using the formula:

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

where;

sample size n = 7

mean = 6+8+5+9+7+5+8 = 48

sample mean $\overline x =\dfrac{48}{7}$

$\overline x$ = 6.86

sample standard deviation is :

$s = \sqrt{\dfrac{\sum( x -\overline x)^2}{n-1}}$

$s = \sqrt{\dfrac{( 6 -6.86)^2+( 8-6.86)^2+ ( 5 -6.86)^2 + ...+( 7 -6.86)^2+ ( 5 -6.86)^2+( 8 -6.86)^2 }{7-1}}$

s = 1.573

population mean $\mu = 3.5$

Therefore, the test statistics is :

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$t = \dfrac{6.86- 3.5}{\dfrac{1.573}{\sqrt{7}}}$

$t = \dfrac{3.36}{\dfrac{1.573}{2.65}}$

$t = \dfrac{3.36\times {2.65}}{{1.573}}$

t = 5.660

At significance level of 0.01

$t_{0.01} = 3.707$

P - value = P(T > 5.66)

P - value = 1 - (T < 5.66)

P - value = 1 - 0.9993

P-value = 0.0007

Therefore, since $t_{0.01} < t$ , we reject the null hypothesis and conclude that the claim that the mean bad debt ratio for Ohio banks is higher than the mean for all financial institutions is true.