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Bad White [126]
3 years ago
13

Estimate the difference of 450.012 and 57.876 to the nearest hundredth. A. 392.1 B. 392.13 C. 392.136 D. 392.14

Mathematics
2 answers:
Doss [256]3 years ago
8 0

Answer:

C

Step-by-step explanation:

Sorry if im wrong i tried my best

OverLord2011 [107]3 years ago
4 0

Answer:

450.012 - 57.876 = 392.136

392.136 rounds to 392.14,so your answer would be D

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adell [148]
X=6
6x5=30
30+10=40
Have a nice day!
5 0
3 years ago
Read 2 more answers
Bus A and Bus B leave the bus depot at 9 am. Bus A takes 30 minutes to complete its route once and bus B takes 40 minutes to com
kirill [66]

Answer:

11 am

Step-by-step explanation:

Bus A and Bus B leave the bus depot at 9 am.

Bus A takes 30 minutes to complete its route once

Bus B takes 40 minutes to complete its route once.

We solve this finding the Lowest Common Multiple of the minutes each bus uses to complete it's route

30 = 3 × 10

40 = 4 × 10

= 3 × 4 × 10

= 120 minutes

120 minutes after 9 am is

60 minutes = 1 hour

60 minutes = 1 hour

= 2 hours.

9am + 2 hours

= 11 am.

Therefore, they be back at the bus depot together at 11 am

4 0
4 years ago
The coins in the store s cash register total $12.50. The cash register contains only nickels, dimes, and quarters. There are twi
Oduvanchick [21]
Nickels is 0,05 of dolar - n
dimes is 0,1 o dolar -d
quarters is 0,25 of dolar -q
dimes =2nickels
d=2n
quarters=2dimes
q=2d=4n
12,5=n(0,05)+2n(0,1)+4n(0,25)
12,5=0,05n+0,2n+n
12,5=1,25n
n=10
q=4n=40
There are 40 quarters
5 0
3 years ago
Question Help Of 515515 samples of seafood purchased from various kinds of food stores in different regions of a country and gen
Setler79 [48]

Answer:

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

Step-by-step explanation:

Sample size, n = 51

p = 0.62

1 - p = 1 - 0.62 = 0.38

n = 515

Confidence level = 90% = Zcritical at 90% = 1.645

Confidence interval = (p ± margin of error)

Margin of Error = Zcritical * sqrt[(p(1-p))/n]

Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]

Margin of Error = 1.645 * 0.0214

Margin of Error = 0.035203

Lower boundary = (0.62 - 0.035203) = 0.584797

Upper boundary = (0.62 + 0.035203) = 0.655203

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

3 0
3 years ago
The length of the human pregnancy is not fixed. It is known that it varies according to a distribution which is roughly normal,
fgiga [73]

Answer:

a) Figure attached

b) P(234

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(234

And we can find this probability with this difference:

P(-2

An we know using the graph in part a that this area correspond to 0.95 or 95%

c) P(250

And we can find this probability with this difference:

P(-1

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

d) P(X

And using the z score we got:

P(X

And that correspond to approximately 0.15%

Step-by-step explanation:

Part a

For this case we can see the figure attached.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part b

Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:

Where \mu=266 and \sigma=16

We are interested on this probability

P(234

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(234

And we can find this probability with this difference:

P(-2

An we know using the graph in part a that this area correspond to 0.95 or 95%

Part c

P(250

And we can find this probability with this difference:

P(-1

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

Part d

We want this probability:

P(X

And using the z score we got:

P(X

And that correspond to approximately 0.15%

5 0
3 years ago
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