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3 years ago

50 points, question is below.

Mathematics

2 answers:

REY [17]3 years ago

8 0

$x^2+3x-10\neq0\\\\x^2+5x-2x-10neq0\\\\x(x-5)-2(x+5)\neq0\\\\(x+5)(x-2)\neq0\to x+5\neq0\ \wedge\ x-2\neq0\\\\x\neq-5\ \wedge\ x\neq2\\\\\dfrac{x+5}{x^2+3x-10}=\dfrac{x+5}{(x+5)(x-2)}=\dfrac{1}{x-2}\\\\Answer:\ x=2$

Naily [24]3 years ago

5 0

Answer:

the answer is X=2

Step-by-step explanation:

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An owner wants to discount his pizza from $15 to $14.25, how much percent does he need to discount?

IrinaVladis [17]

Answer:

Step-by-step explanation:

Amount of money that is discounted

= $15 - $14.25

$0.75

Percentage that needed to be discount

=(0.75/15) * 100%

= 5%

Thus, owner needs to discount 5% to discount his pizza from $15 to $14.25.

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3 years ago

Solid: A square pyramid. The square base has side lengths of 5. The 4 triangular sides have a base of 5 and height of 7. What is

lina2011 [118]

Answer:

95 square units

Step-by-step explanation:

Surface area of pyramid = 4(side area) + base

Solve for the base first.

Area of base = s*s = 5*5 = 25

Next solve for one of the sides.

Area of triangle = 1/2 b*h = 1/2 5*7 = 17.5

Plug these back into our original equation.

Surface area of pyramid = 4(side area) + base

Surface area of pyramid = 4(17.5) + 25

Surface area of pyramid = 70 + 25 = 95 square units

4 0

4 years ago

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dolphi86 [110]

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3 years ago

2-3. The triangles below are drawn to scale and congruent sides and angles are indicated with tick marks.

mihalych1998 [28]

Answer: The triangles are congruent by SAS(side-angle-side)

Step-by-step explanation:

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6 0

3 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

4 years ago