What was John’s age 4 years ago, if he will be z years old in 5 years?

Mathematics

1 answer:

enot [183]3 years ago

5 0

Answer:

z-9

Step-by-step explanation:

Right now, John is z-5. z-5 - 4 is equal to z -9.

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Step-by-step explanation:

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1 year ago

Replace the ? with one of the following symbols (<, >, =, or ≠) for 4 + 3 + 7 ? 7 + 0 +7 A. > B. ≠ C. = D. <

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There is a strong positive linear correlation between rainfall and the number of oranges a tree produces. Does this mean that mo

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Read 2 more answers

Choose ALL possible answers

muminat

Select all the correct answers:
1) Yes

2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)

3) Yes

4) No

5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20

rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18

rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15

rg=20>18=rf
rg=20>15=rh

6) No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)

7 0

2 years ago

The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds

Bumek [7]

Answer: 11.5%

Explanation:

Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds.

Now, we get the z-score for 720 seconds by the following formula:

$\text{z-score} = \frac{x - \mu}{\sigma}$

where

$t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}$

So, the z-score of 720 seconds is given by:

$\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}$

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

$P(t \ \textgreater \ 720) 
\\ = P(z \ \textgreater \ 1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\ \boxed{P(t \ \textgreater \ 720) = 0.115}$

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice.

8 0

2 years ago