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Ad libitum [116K]
3 years ago
12

Work out these questions without a calculator and showing all working out

Mathematics
1 answer:
natima [27]3 years ago
3 0
1) 6÷0.2 = 30

If 6/2=3 then 6/0.2=30 as the decimal place shifts one place.

2)8÷0.1 = 80

8/1=8 so shift the decimal place over once to make 80.

3)9÷0.3 = 30

9/3=3 so shift the decimal place over once to get 30.

4)4÷0.04 = 100

4/4=1 so shift the decimal place over twice to get 100.

5)7÷0.002 = 3500

7/2=3.5 so shift the decimal place over three times to get 3500

6)0.718÷0.2 = 3.59

718/2=359 so shift the decimal over three places for the 0.718 and then back over once for the 0.2

7)0.0141÷0.003 = 4.7

141/3=47 so shift the decimal over our times for the 0.0141 and then back over three times for the 0.003

8)0.24÷0.012 = 20

24/12=2 so shift the decimal point over once twice for 0.24 then back over three times for 0.012

9)1.625÷0.0013 = 1250

1625/13=125 so shift the decimal point over three times for the 1.625 and then back four times for the 0.0013

10)47.1÷0.15 = 314

471/15=31.4 so shift the decimal point over once for the 47.1 and then back over twice for the 0.15.

Hope this helps :)
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The distance from first base to third base is 1.4times the distance from home plate to third base, which is k feet. REPRESENT TH
geniusboy [140]

Answer:

1.4 times k

Step-by-step explanation:

If k feet is the distance from home plate to third base, and the distance from first base to third base is 1.4 times the previous value, then

distance from first base to third base = 1.4 * k or 1.4 times k

If, for example, k=90 feet, then from first to third base there are

1.4*90=126 feet

4 0
3 years ago
Tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the e
ioda

Answer:

Step-by-step explanation:

Given that tags are  attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the event that the right leg tag is lost. Suppose those two events are independent and A2 the event that the right leg tag is lost. Suppose those two events are independent and P(A1) = P(A2) = 0.3

a)  the probability that at least one leg tag is lost

= P(A_1 U A_2)\\= P(A_1) + P(A_2)-{(A_1 \bigcap A_2)\\=P(A_1) + P(A_2)-{(A_1 )P(A_2)

(since A1 and A2 are independent)

=0.3+0.3-0.09

=0.51

b) the probability that exactly one tag is lost, given that at least one tag is lost.

= P(one leg lost)/P(atleast one leg lost)

=P(atleast one leg lost)-P(Both lost)/0.51\\= \frac{0.51-0.09}{0.51} \\=14/17

c) the probability that exactly one tag is lost, given that at most one tag is lost.

4 0
3 years ago
on the last friday in may, one fourth of the 280 students in a school were away on a field trip how many students were on the fi
Stells [14]
Seventy that’s just simple math lol. Your welcome and hopes it helped
6 0
3 years ago
How would you rewrite -9/-6
attashe74 [19]

Answer:

3/2 or 1.5

Step-by-step explanation:

A negative divided by a negative makes the quotient a positive, 9 divided by 6 is 1.5 and the gcf of 9 and 6 is 3, so you would simplify it by doing 9 divided by 3 and 6 divided by3 which is 3/2

4 0
3 years ago
5x − 3y − z if x = −2, y = 2, and z = −3
Marina CMI [18]
Just plug in and solve 
5x − 3y − z → <span>5(-2) − 3(2) − (-3)

</span>5(-2) − 3(2) − (-3) → -10 - 6 +3

-10 - 6 +3 → -16 + 3 

-16 + 3 = -13 

answer is -13


7 0
4 years ago
Read 2 more answers
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