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3 years ago

True or false pls help me

Mathematics

2 answers:

worty [1.4K]3 years ago

8 0

Answer:

B) False.

Step-by-step explanation:

Really

stellarik [79]3 years ago

5 0

Answer:

False

Step-by-step explanation:

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Bargain Feed Store has a markup rate of 14.1%. What price will the shop charge for a bag of horse feed that cost $12.22 wholesal

ratelena [41]

Answer: $13.94

Step-by-step explanation:

Given : Bargain Feed Store has a markup rate of 14.1%.

The wholesale cost of bag of horse feed = $12.22

Price mark up = Mark up % x Whole sale cost

Price mark up =14.1% x $12.22

Price mark up $=\dfrac{14.1}{100}\times\$12.22=\$1.72302\approx\$1.72$

Then , the final price of bag of horse feed will be :-

Final price = wholesale cost +Price mark up

=$12.22+$1.72 = $13.94

Hence, the shop will charge $13.94 for a bag of horse feed that cost $12.22 wholesale.

6 0

2 years ago

What is the value of x

Ainat [17]

Answer:

Step-by-step explanation:

(3x +3) + (5x-2) + (3x+3) =180

3x +3 + 5x-2 + 3x+3 =180

11x +4=180

11x =180-4

11x=176

x=176/11

x= 16

4 0

2 years ago

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet

frosja888 [35]

There seems to be one character missing. But I gather that f(x) needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume f(x) is monic, meaning the coefficient of the leading term is 1, or

$f(x) = x^5 + \cdots$

Since $x^3$ divides $f(x)$ , and

$f(x) = x^3 p(x)$

where $p(x)$ is degree-2, and we can write it as

$f(x)=x^3 (x^2+ax+b)$

Now, we have

$f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2$

so if $(x - 1)^3$ divides $f(x)^2$ , then $p(x)$ is degree-2, so $p(x)^2$ is degree-4, and we can write

$p(x)^2 = (x-1)^3 q(x)$

where $q(x)$ is degree-1.

Expanding the left side gives

$p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2$

and dividing by $(x-1)^3$ leaves no remainder. If we actually compute the quotient, we wind up with

$\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}$

If the remainder is supposed to be zero, then

$\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}$

Adding these equations together and grouping terms, we get

$(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1$

Then $b=-1-a$ , and you can solve for a and b by substituting this into any of the three equations above. For instance,

$2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1$

So, we end up with

$p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}$

6 0

2 years ago

Helpppp me pleaaaaseeeewww

s2008m [1.1K]

Answer:

Step-by-step explanation:

I believe it is A If I remember correctly

8 0

3 years ago

In a multiple linear regression analysis, statistical model utility is determined by the value of R2 and 2s, while practical mod

Natali [406]

Answer:

This statement is true because if F statistics is significant, then the entire multiple regression model is useful for the prediction of y.

Step-by-step explanation:

Solution:

In a multiple linear regression analysis, the statistical model utility is determined by the value of R2 and ∂2, while the Global F test p-value determines the practical model.

This statement is true because if F statistics is significant, then the entire multiple regression model is useful for the prediction of y.

The F test is a statistical test that is very flexible. F test can evaluate multiple model terms.

The F test indicates whether a linear regression model provides a better fit to data than a model that contains no independent variable.

The F test for overall significance has two hypotheses:

1- Null hypothesis:

The fit of the intercept only model and your model are equal.

2- Alternative hypothesis:

The fit of the intercept-only model is significantly reduced compared to your model.

Compare p-value for the F test,

If the p-value is less than the significance level, the sample data provide sufficient evidence that the regression model fits the data better with no independent variable.

5 0

2 years ago