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2 years ago

Julian fully simplifies this polynomial and then writes it in standard form.

Mathematics

1 answer:

Scilla [17]2 years ago

7 0

Answer:

4y^4

Step-by-step explanation:

$4x^2y^2 - 2y^4 - 8xy^3 + 9x^3y + 6y^4 - 2xy^3 - 3x^4 + x^2y^2$

Last term is -3x^4

This means x term is increasing and it ends at x^4

we arrange the terms in order

$6y^4- 2y^4 - 8xy^3 - 2xy^3 +4x^2y^2+ x^2y^2 + 9x^3y-3x^4$

now we combine like terms

$4y^4 - 10xy^3 +5x^2y^2 + 9x^3y -3x^4$

So first term is 4y^4

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Aarti bought a square shaped table cloth for her home the side of the table cloth measures 2 1/3m what is the area of the table

PSYCHO15rus [73]

Answer:

Area of table cloth = 49/9 m² or 5.44 m²

Step-by-step explanation:

Given:

Side of table cloth = $2\frac{1}{3}m$ = 7/3 m

Shape of cloth is square

Find:

Area of table cloth

Computation:

Area of square = side²

So,

Area of table cloth = side²

Area of table cloth = (7/3)²

Area of table cloth = 49/9 m² or 5.44 m²

3 0

1 year ago

6.In a sample of 131 women with cosmetic dermatitis from using eye shadow, 12 were diagnosed with a nickel allergy. In a sample

aivan3 [116]

Answer:

a) $0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

b) $\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

c) For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

Step-by-step explanation:

Part a

The estimated proportion of women with cosmetic dermatitis from using eye shadow is given by:

$\hat p =\frac{12}{131}= 0.0916$

The confidence interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the proportion is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing we got:

$0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

Part b: A 95% confidence interval for the women with cosmetic dermatitis from using mascara

$\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

Part c: Suppose you are informed that the true proportion with a nickel allergy for one of the two groups (eye shadow or mascara) is .12. Can you determine which group is referenced? Explain.

For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

7 0

2 years ago

Bill is considering taking out a loan. He estimates that he can afford monthly payments of $375 for 10 years in order to support

Anuta_ua [19.1K]

For the same payment and term, the lower the interest rate, the more the loan can be. The opposite is also true. The appropriate selection is ...
B. If the interest rate were 6.8%, the amount of the loan Bill is considering taking out would be less than $33,025.69.

_____
At 6.8%, Bill can afford a loan of $32,585.93.

4 0

2 years ago

6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability th

Maru [420]

Answer:

(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.

(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

Step-by-step explanation:

Let X = number of heads.

The probability that a head occurs in a toss of a coin is, p = 0.50.

The coin was tossed n = 100 times.

A random toss's result is independent of the other tosses.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.50.

But the sample selected is too large and the probability of success is 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of $\hat p$  (sample proportion of X) if the following conditions are satisfied: