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Sedbober [7]
3 years ago
11

i used the answer key to get my answer and checked it with my dumb useless teacher and it’s right but, he won’t tell me how to g

et the numerators like how does it go from 3 & 7 to 15 & 14 please help omg!!!!

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
You multiply it with the same number you use for the denominator.
in this case 3/4 and 7/10
you find the LCM of the denominator is 20
so for 4 to become 20 you multiply by 5 then for the numerator 3 also multiply by 5, so you get 15/20
its the same also with 7/10, you multiply both the numerator and the denominator by 2 and get 14/20
hope it help... :)




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Orlov [11]

Answer:

your answer will be 11

Step-by-step explanation:

Thanks again!!

4 0
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Use the relationship between the angles in the figure to answer the question.
Brut [27]

Answer:

​x + 49 + 28 = 180​

Step-by-step explanation:

​The angles on a straight line with x° are 28° and 49° respectively, side by side with x°. This is because, vertical angles are equal. Therefore, the angle vertically opposite 28° and 49° are equal to 28° and 49° respectively.

Therefore, since angles on a straight line is 180°, thus:

​x + 49 + 28 = 180​

6 0
3 years ago
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A water tower that is 132 feet tall casts a 48-foot long shadow. If Shannon is 1 point
MariettaO [177]

Answer:

2.03 ft

Step-by-step explanation:

132 divided by 48 is 2.75 and 5.6 divided by 2.75 is 2.03

6 0
3 years ago
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
Six groups of students sell 162 balloons at the school carnival. there are 3 students in each group. if each student sells the s
Marina CMI [18]
Since there are six groups of student and each group consists of 3 students, therefore, the total number of students is 18. All of them sold 162 balloons. In order to know how many balloons each student sold, we just have to divide 162 by 18 and the answer is 9. Hope this answers your question.
4 0
3 years ago
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