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2 years ago

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In Westchester County, the average home sells for $750,000. A survey of 250 applicants found that 38% of homebuyers are willing

to spend $875,000 on a single family home. Identify the sample size

2 answers:

miskamm [114]2 years ago

7 0

Answer: 250
explanation:

Oliga [24]2 years ago

6 0

Answer is 250 good luck

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marin [14]

Answer:

12

Step-by-step explanation: So we know the car has 8 gallons of gas. 1/3 would be 4 gallons. 2/3 would be 8 gallons. 12 gallons would be 3/3. my brain isnt working rn but i used mt last braincell umm i suppose its correct

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2 years ago

There were 7 inches of rain in 24 hours. What is the average amount of rain per hour

Mademuasel [1]

Answer:

7 /24 inch per hour

Step-by-step explanation:

To find the average amount of rain, take the total amount of rain and divide by the time

7 inches / 24 hours

7 /24 inch per hour

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2 years ago

A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular store that have an a

storchak [24]

Answer: 267.

Step-by-step explanation:

When there is no prior information for the population proportion, then the formula we use to find the sample size to estimate the confidence interval :

$n=0.25(\dfrac{z^*}{E})^2$ , where z* = Critical z-value and E + amrgin of error.

Let p = proportion of packages of ground beef sold at a particular store that have an actual fat content exceeding the fat content stated on the label.

Since , we have no prior information about p. so we use above formula

with E = 0.06 and critical value for 95% confidence =z* =1.96 [By z-table ] , we get

$n=0.25(\dfrac{1.96}{0.06})^2=0.25(32.6666)^2\\\\=(0.25)(1067.111111)\\\\=266.777777778\approx267$

Hence, the required sample size is 267.

7 0

2 years ago

20 to 8 in simpliest form

Mashcka [7]

Answer:

in it's lowest form its 5/2

8 0

2 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

2 years ago