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s344n2d4d5 [400]
3 years ago
10

Sujin babysat 20 hours this week. That is 4 times as long as she babysat last week. How long did she babysit last week?

Mathematics
2 answers:
inessss [21]3 years ago
6 0
Five hours

explanation:
Divide 20 by 4 to find your answer
jeyben [28]3 years ago
4 0

Answer:

5 hours

Step-by-step explanation:

let the time she baby sat be x.

4x= 20

x= 5

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What is the arithmetic sequence of a1=228 n=28 sn=2982
Art [367]

Answer:

use the formula sn= n(a1+an)/2

Step-by-step explanation:

2982=28(228+an)/2

5964=28(228+an)

5964/28=228+an

213=228+an

an=-15(last term)

to find difference use formula

an = a+(n-1)d

-15=228+(28-1)d

-243=27d

d=-243/27

d=-9

arithmetic sequence  can be found be keep on subtracting 9 from 228

hence the arithmetic sequence is

228, 219, 210, 201, 192, 183, 174........-15

3 0
3 years ago
A group of cubes are arranged in a pattern with the same number of cubes in each row and column. There are 225 cubes in the shap
Maslowich

Answer:

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3 0
4 years ago
Problem: Suppose that on-the-job injuries in textile mill occur at the rate of 0.1 per day. What is the probability that two acc
sertanlavr [38]

Answer: the probability that two accidents will occur during the (five-day) work week is 0.0758

Step-by-step explanation:

Given that;

injuries in textile mill occur at the rate of 0.1 per day; λ = 0.1

probability that two accidents will occur during the (five-day) work week = ?

T= 5

so, the average number of on the job injuries for five days will be;

λT = 0.1 × 5

= 0.5

now if X follows Poisson distribution, hemce

P( X=k ) = [e^-λT (λT)^k] / k!

so the probability that two (2) accidents will occur during the (five-day) work week will be;

P( X=k ) = [e^-0.5 (0.5)²] / 2!

= (0.6065 × 0.25) / 2

= 0.1516/ 2

= 0.0758

the probability that two accidents will occur during the (five-day) work week is 0.0758

3 0
3 years ago
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