The weight of a pig is 98 pounds to the nearest half of a pound. What is the maximum error of this measurement?

Mathematics

1 answer:

Orlov [11]3 years ago

4 0

Answer:

1/4 pound

Step-by-step explanation:

The maximum error is half the "rounded to" value. Here, rounding is to 1/2 pound, so the maximum error is 1/4 pound.

You might be interested in

How do I solve 5 + 6y = 9y + 2 + 3(1 - y)

Stella [2.4K]

Actually you can't solve this. There is no solutions.

3 0

3 years ago

Read 2 more answers

here is a list of numbers: 12, 13, 19, 16 ,32, 15, 13.a)work out the range of the numbers in the list.b)work out the mean of the

ratelena [41]

Answer: Mean: 17.14 Range: 20.

Step-by-step explanation: 12+13+13+15+16+19+32 25+13+15+16+19+32 38+15+16+19+32 53+16+19+32 69+19+32 88+32 120/7 17.14

32-12 20

6 0

3 years ago

Why and when do we cross multiply in algebra?

Nataly [62]

Answer:

You cross multiply to solve for a variable. You cross multiply when given a proportion.

4 0

3 years ago

Read 2 more answers

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$