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galben [10]
2 years ago
13

Julio is selecting a random sample of people to survey for a newspaper article. Which elements are important for Julio to consid

er when choosing a random sample? Select three options.
the size of the population
the size of the sample
the characteristics and demographics of the population
the probable responses to the survey questions from the population
the probable responses to the survey questions from the sample
Mathematics
2 answers:
Keith_Richards [23]2 years ago
7 0

Answer:

A, B, and C

Step-by-step explanation:

Fofino [41]2 years ago
5 0

Answer:

Step-by-step explanation:

Answer:

It say the size of population characteristics

Step-by-step explanation:

Because  it makes sense those are I need

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A train travelled 275 km in 2.5 hours how fast will the train go in an hour?
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Carla wants to start a college fund for her daughter Lila. She puts $63,000 into an account that grows at a rate of 2.55% per ye
Oksanka [162]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 63000

r = 2.55% = 2.55/100 = 0.0255

n = 12 because it was compounded 12 times in a year.

Therefore, function, C(t), that represents the amount of money in the account t years after the account is opened is

C(t) = 63000(1 + 0.0255/12)^12t

C(t) = 63000(1.002125)^12t

For C(t) = 100000,

100000 = 63000(1.002125)^12t

100000/63000 = (1.002125)^12t

1.587 = 1.002125)^12t

Taking log of both sides

Log 1.587 = log 1.002125)^12t

Log 1.587 = 12tlog 1.002125)^

0.2005 = 12t × 0.00092

0.2005 = t × 0.01104

t = 0.2005/0.01104

t = 18.16 years

5 0
3 years ago
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams
Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0
3 years ago
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