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Furkat [3]
3 years ago
10

(20 PTS!!)

Mathematics
2 answers:
creativ13 [48]3 years ago
7 0

Answer:

The correct option is C.

Step-by-step explanation:

In a dot plot, the number of dots above a number represents the frequency of that number.

From the given dot plot it is clear that number of dots for group R is 21 and the time taken by the students of groups R are

45, 50, 50, 55, 60, 60, 70, 75, 80, 85, 90, 90, 95, 100, 100, 105, 110, 110, 115, 120, 120

The average of the data is

Average=\frac{\sum x}{n}

Average=\frac{45+50+50+55+60+60+70+75+80+85+90+90+95+100+100+105+110+110+115+120+120}{21}

Average=\frac{1785}{21}

Average=85

From the given dot plot it is clear that number of dots for group S is 17 and the time taken by the students of groups S are

50, 55, 60, 60, 65, 70, 70, 80, 90, 90, 95, 100, 105, 110, 110, 115, 120

Average=\frac{50+55+60+60+65+70+70+80+90+90+95+100+105+110+110+115+120}{17}

Average=\frac{1445}{17}

Average=85

The average time of both groups are same.

Since both groups show about the same average time, therefore the correct option is C.

adell [148]3 years ago
4 0
The answer would be "both groups show about the same average time"
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Toward the middle of the harvesting season, peaches for canning come in three types, early, late, and extra late, depending on t
Goryan [66]

Answer:

a) 12

b) 18

c) 37

d) 97

Step-by-step explanation:

Toward the middle of the harvesting season, peaches for canning come in three types, early, late, and extra late, depending on the expected date of ripening. During a certain week, the following data were recorded at a fruit delivery station:

34 trucks went out carrying early peaches ;

61 carried late peaches ;

50 carried extra late;

25 carried early and late;

30 carried late and extra late;

8 carried early and extra late;

6 carried all three;

9 carried only figs (no peaches at all) .

<u>To solve this problem, the best solution is to drawn a Venn Diagram, which is attached.</u>

(a) How many trucks carried only late variety peaches?

12

(b) How many carried only extra late?

18

(c) How many carried only one type of peach ?

7+12+18 = 37

(d) How many trucks (in all) went out during the week?

Sum of all the numbers on the Diagram:

7+2+6+19+12+24+18+9= 97

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3 0
4 years ago
Read 2 more answers
1. The monthly rents for studio apartments in a certain city have a mean of $920 and a standard deviation of $190. Random sample
Veronika [31]

Answer:

1)Mean=920 $

2)Standard Error =S.E.=38

3)more likely to take place Event 25 studio apartments with a mean rent between $901 and $939

Step-by-step explanation:

Given:

True mean =920 $

S.D=190 $

No .of samples=25

To Find:

1)Mean of sample.

2)S.E

Solution:

Now , the mean for given sample distribution is  given as 920 $

Hence Mean =920 $

Now calculating the Standard error

S.E.= S.D./Sqrt(n)

=190/Sqrt(25)

=190/5

=38

Therefore the Standard error is about 38 .

a) When n=1 then P(901≤X≤939)

So,

Z1=(901 -920)/190/Sqrt(1)]

Z1=-0.1

Z2=(939-920)/(190/Sqrt(1)]

Z2=0.1

So

Pr(-0.1≤Z≤0.1)=P(Z≤0.1)-P(Z≤-0.1)

=0.5398-0.4602

=0.0797  % chance of the sample distribution

b)n=25 then P(901≤X≤939)

Z1=(901-920)/S.E

Z1=-19/38=

Z1=-0.5

And Z2=0.5

So,

Pr(-0.5≤Z≤0.5)

=Pr(Z≤0.5)-Pr(Z≤-0.5)

=0.6915-0.3085

=0.3829

i.e 38.29 % chance of the sample distribution

Hence More likely to take place will be a sample of 25 studio apartments with a mean rent between $901 and $939.

0.3829>0.0797.

Because the probability of causing above event is more than the option A.

3 0
3 years ago
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