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serg [7]
3 years ago
12

A water balloon is 5 feet above the ground when Sally launches it into the air. Use the quadratic equation 0 = -t2 + 4t + 5 to f

ind how much time, t, it takes for the water balloon to reach the ground.
Mathematics
1 answer:
Alborosie3 years ago
6 0

Answer:

2-sqrt14/2, 2+sqrt13/2.

Step-by-step explanation:

What you do is you have to do the quadratic equation like it says in the problem.

x=  −b± sqrtb^2 −4ac /2a .

a=-2, b=4, c=5.

x=-4±sqrt(4)^2-4(-2)(5)/2(-2).

x=-4±sqrt16+40/-4.

x=-4±2sqrt14/-2.

2-sqrt14/2, 2+sqrt13/2. is your answer once you have done everything.

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70 + x + 55 + 30 = 180
sergiy2304 [10]

Answer:

x=25

Step-by-step explanation:

First, we add all like terms. Then, we solve for x.

70+x + 55 + 30 = 180\\x+70 + 55 + 30 = 180\\x+155=180\\x=25

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2 years ago
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18 POINTS will rate brainiest 6th grade math
stira [4]

Answer:

18 ft^2

Step-by-step explanation:

The area of a parallelogram is A=bh.

The area would be A = 9*4.

A = 36

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36/2 = 18 ft^2

5 0
2 years ago
Is education related to programming preference when watching TV? From a poll of 80 television viewers, the following data have b
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
3 years ago
Can someone help me solve this and explain it to me? <br> −1/6(x−18)+1/3(x+3)=x+4
KATRIN_1 [288]

Answer:

Below.

Step-by-step explanation:

So calculate.

-1/6x+3+1/3x+1=x+4.

Cancel equal terms.

-1/6x+4+1/3x=x+4

Multiply both sides -1/6x+1/3x=x

collect like terms.

-x+2x=6x

move the variable from right to left.

x=6x

collect like terms

x-6x=0

Divide both sides

-5x=0

x=0

7 0
2 years ago
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