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dimaraw [331]
3 years ago
7

Look at the image please helppp

Mathematics
1 answer:
balu736 [363]3 years ago
7 0

Answer:

Option A. one rectangle and two triangles

Option E. one triangle and one trapezoid

Step-by-step explanation:

step 1

we know that

The area of the polygon can be decomposed into one rectangle and two triangles

see the attached figure N 1

therefore

Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles

so

A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2

step 2

we know that

The area of the polygon can be decomposed into one triangle and one trapezoid

see the attached figure N 2

therefore

Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid

so

A=\frac{1}{2}(8)(4)+\frac{1}{2}((4+8)(8)=16+48=64\ yd^2

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The Math Club is baking pies for a bake sale. The fruit-pie recipe calls for twice as many peaches as nectarines. If it takes a
Natali5045456 [20]

Answer:

56 number of  Nectarines are needed

Step-by-step explanation:

Let  number of Fruit-pie be x and  number of nectarines be y.

Since the fruit pie recipe calls for twice as many peaches as nectarines.

then we have,

x= 2y

It is also given that total of 168 pieces fruit for all pies.

⇒ x+y=168

Substitute x=2y in above equation to solve for y:

 2y+ y =168 or

3y=168

Divide both sides by 3, we get

\frac{3y}{3}= \frac{168}{3}

Simplify:

y = 56

Therefore, 56 nectarines are needed.


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3 years ago
Find the slope of the line. -2,-3 1,3
Serhud [2]

Answer:

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3 years ago
In a play-school with 500 students, the number of boys is 20 more than five-sevenths of the number of girls. Let’s find the numb
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Step-by-step explanation:

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3 years ago
A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
KiRa [710]

Answer:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

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:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

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

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

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}

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{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

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

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

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

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

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