How do I solve 1 1/3 × 2 2/4 is steps?

How are polynomials like and unlike monomials? Provide an example of a monomial and an example of a polynomial. How does the fol

Ainat [17]

A polynomial is the sum of at least one term. For example, x^3+1 is a polynomial. A monomial is a polynomial with only one term, such as 2x^2. A binomial is a polynomial with two terms, and a trinomial is one with three terms. The example you gave is a trinomial (which is also a polynomial). Degree of a polynomial is the largest sum of variable powers in any term of the polynomial. So, for example, x^2 y has degree 3, and x^3+x^2 also has degree 3. A sixth degree polynomial would be x^6-2x+1, for example.

8 0

3 years ago

20. Solve the equation x2 - 17x + 72 = 0.

Rina8888 [55]

Answer:

x=8,9

Step-by-step explanation:

x^2-17x+72=0

(x-8)(x-9)=0

x=8,9

3 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago

I don't know which one is the correct answer? PLEASE HELP!!

alukav5142 [94]

The answer is 1/q^30 because after simplifying, you can cancel out the p and then you get q^q-20 times q^-10 to get q^-30 which then equals 1/q^30

7 0

2 years ago

Read 2 more answers

|x + 5| – |2w| When x = 2 and w = –4

Vitek1552 [10]

15
(2 + 5) - (2 x -4) = 15
7 - -8
Minus a negative goes upwards.
15

4 0