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Misha Larkins [42]
3 years ago
15

Determine whether the situation involves permutations or combinations. Then solve. How many ways are there to place an algebra b

ook, a geometry book, a chemistry book, an English book, and a health book on a shelf?
Mathematics
1 answer:
Nikitich [7]3 years ago
5 0

Answer:

<h2>The situation involves permutation</h2><h2>The number of arrangement is 120</h2>

Step-by-step explanation:

Given that

Algebra book=1  

Geometry book=1

Chemistry book= 1

English book= 1

Health book= 1

Total number of books N = (1+1+1+1+1)= 5

      Permutation is used to  determines the number of possible arrangements in a set when the order of the arrangements is crucial.

Number of arrangements = N!

Number of arrangements= 5*4*3*2*1=  120

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The third one tells us the price is $12 per liter, that is to say, the top number is twelve times the bottom number.

.25 × 12 = $3

.7 × 12 = $8.40

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3.52 × 12 = $42.24

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2 years ago
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Rina8888 [55]

Answer:

For the first question, the answer is rotation 290 counterclockwise.

And for the second question, the answer is accurate.

8 0
3 years ago
A total of 765 tickets were sold for the school play. They were either adults or students tickets. There were 65 more student ti
Brrunno [24]

Answer: 350 adult tickets

Step-by-step explanation:

(omg I remember this question!)

  • a stands for the number of adult tickets sold

student tickets : a + 65

<em>the equation for the prob: </em>

765 = a + (a + 65)

<em>solve:</em>

combine 'like terms'

1.) 765 = a + a + 65

2.) 765 = 2a + 65

<u>- 65         - 65 </u>

700= 2a

divide by 2

700/2 = 2a/2

<em>(700/2 = 350) </em>

<em>(the "2" in 2a is cancelled out by the other 2)</em>

<u>350 = a </u>

7 0
2 years ago
Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct resul
o-na [289]

Answer:

H0:p= 0.80                 H1: p< 0.80  one tailed test

Step-by-step explanation:

We state the null and alternative hypotheses as that the results are 80 % against the claim that the results are less than 80%.

H0:p= 0.80                 H1: p< 0.80  one tailed test

p2= 0.8 , p1= 74/97= 0.763

q1= 1-0.763= 0.237    q2= 0.2

The level of significance is 0.05 .

The Z∝= ±1.645 for ∝= 0.05

The test statistic used here is

Z= p1-p2/ √pq/n

Putting the values:

Z= 0.763 -0.8 / √ 0.8*0.2/97

z= -0.037/ 0.0406

z= -0.9113

The Z∝ = ±1.645 for ∝= 0.05 for one tailed test.

As the calculated value  does not fall in the critical region  we fail to reject the null hypothesis. There is not sufficient evidence to support  the claim that such polygraph results are correct less than 80% of the time.

Using the normal probability table.

P (Z <  -0.9113)= 1- P(z= 0.9311) = 1- 0.8238= 0.1762

If P- value is smaller than the significance level reject H0.

0.1762> 0.005  Fail to reject H0.

8 0
3 years ago
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