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kipiarov [429]
3 years ago
11

Divide.Yes Rectangular models to record the partial quotient for

Mathematics
1 answer:
artcher [175]3 years ago
5 0

Answer:

a. To find 246 divided by 3 by partial quotients .

Here dividend= 246 and divisor =3

Step 1: Write easy multiples of divisor that is nearest to the dividend as follows

3×10=30

3×20=60

3×30=90

....3×80=240(nearest easy factor)

Step 2: Now subtract 240 from 246 ,we get 6 then subtract 6=3×2 from it we get 0.

Step 3: <em>Adding the partial quotients</em>=80+2=82

i.e. 246=3(80)+3(2).....> Quotient

b. 126÷2

Here dividend= 126 and divisor =2

Write easy multiple

2×10=20

2×20=40

....3×60=120(nearest easy factor)

Now subtract 120 from 126 ,we get 6 then subtract 6=2×3 from it we get 0.

<em>Adding the partial quotients</em>=60+3=63.....> Quotient

c. . 605÷5

dividend= 605 and divisor =5

Write easy multiples

5×10=50

5×20=100

5×30=150

....5×120=600(nearest easy factor)

Step 2: Now subtract 600 from 605 ,we get 5 then subtract 5=5×1 from it we get 0.

Step 3: <em>Adding the partial quotients</em>=120+1=121.....> Quotient



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Answer:

-6\leq x\leq 16

Step-by-step explanation:

So we have the inequality:

|x-5|\leq 11

Definition of Absolute Value:

x-5\leq 11\text{ or } x-5\geq -11

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x\leq 16\text{ or } x\geq -6

Merge:

-6\leq x\leq 16

And we're done!

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3 years ago
Suppose that each observation in a random sample of 100 fatal bicycle accidents in 2015 was classified according to the day of t
liraira [26]

Answer:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0:  proportion of fatal bicycle accidents in 2015 was the same for all days of the week

against the claim

Ha:  proportion of fatal bicycle accidents in 2015 was not the same for all days of the week

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (ni - npi)²/ npi

which has an approximate chi square distribution with ( n-1)=7-1=  6 d.f

4) The critical region is χ² ≥ χ² (0.05)6 = 12.59

5) Calculations:

χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28

χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]

χ²= 1/14.28[8.1364]

χ²= 0.569= 0.57

6) Conclusion:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week

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